tisdag 19 maj 2015

edX: "Back Radiation" as the Physics of the "Greenhouse Effect"

I started my journey as climate skeptic in 2009 in an attempt to understand the physics of the so called "greenhouse effect" threatening human civilisation by global warming from human emission of CO2 as a powerful "greenhouse gas".

I then discovered that the "greenhouse effect"was (and still is) is very vaguely identified in the scientific literature which poses a severe difficulty to skepticism of CO2 alarmism.

The ongoing edX course Denial101x Making Sense of Climate Science Denial is an attack on skepticism to CO2 alarmism, referred to as "denialism", introduced by:
  • In this first week you will be introduced to some of the terminology we will use in the course in order to begin building your understanding of scientific consensus, the psychology of denial and the spread of denial. 
The course shows which skeptics arguments are the most effective and thus requires special effort to kill. In this sense the course offers valuable insight. In a video lecture in week 3 on the "Greenhouse Effect", we are informed that:
  • The glow from the Earth surface goes upwards, greenhouse gasses absorb some of this heat and they then glow in every direction including down towards us. 
  • This is how the greenhouse effect works. We measure it every day here at Reading Atmospheric Obervatory by a pyrgeometer...it has a special window only allowing infrared light through to be measured. Even during a cloudless night it measures the constant greenhouse glow. 
  • Even though the greenhouse effect is an observed fact, there is a myth that it does not exist. This myth misinterprets a law of physics called the second law of thermodynamics. The 2nd law says that even though heat moves in all directions, overall heat moves from hot to cold, and not from cold to hot. 
  • The myth says that the greenhouse effect does not exist because it means heat moving from a cooler sky to warmer surface. But this is a misrepresentation: The greenhouse effect obeys the law: A square meter of Earth surface send about 500 Watts upwards, so it works like a 500 W heater. The greenhouse effect sends down about 330 Watts of heat, so in total about 170 Watts goes from the warmer surface to the cooler sky. Heat overall goes from hot to cold  but the greenhouse effect sends som back to warm us up.
  • The myth misrepresents the 2nd law. Meanwhile observatories measures the greenhouse effect every day all over the globe.
We understand that the "greenhouse effect" is based on "back radiation" from the atmosphere, which is measured by pyrgeometers. 

From the beginning of my skeptics journey I understood, from a new proof of Planck's radiation I had constructed as part of a larger effort to describe physics as analog computation,  that "back radiation" is an illusion without physical reality, and so that a pyrgeometer is constructed to sell this illusion to a market in need of "instrumental evidence".  

This insight has made me into a "denier" in the view of not only alarmists but strangely enough also in the view of leading skeptics such as Singer and Spencer and many others.  As a "denier of back radiation" based on a view of physics as computation,  I have met many strong reactions often including direct censorship of this my view. 

The edX course gives me more courage to not give up this view including a new proof of Planck's law leading to the conclusion that "back radiation" is non-physical illusion. The edX course shows that if "back radiation" is illusion, then so is the "greenhouse effect". Unfortunately, leading skeptics have fallen into the trap of "How to fool yourself with a pyrgeometer".

You find more material under the categories "myth of back radiation" and "pyrgeometer". To get rid of illusions may get very quickly, once you meet the right argument. Notice in particular the recent post on the unphysical aspect of Schwarzschild's radiation model introducing the unfortunate unphysical idea of "back radiation" or "downwelling longwave radiation DLR" as the warming element of the "greenhouse effect".

Take a special look at the argument presented: The 2nd law says that even though heat moves in all directions, overall heat moves from hot to cold.  If anything, this is a false version of the 2nd law: It is not true that "heat moves in all directions".

I asked edX to give the scientific justification of this statement, and report the answer. If you go through the response from edX below, you will find that after an exchange of nearly 100 comments back and forth, we are still far from getting an answer from edX. The tactic used by edX is to meet any question from me as a student following the course, by a battery of counter-questions with the objective of keeping me busy and so avoiding to answer my question. Clever, but tiresome both for edX and me.

Here is a copy of the edX Discussion with teacher Gavin Cawley:

Gavin: The physical mechanism is very straightforward. I would happily go through the physics with you, step-by-step, starting with back body radiation, to see where we agree and where we disagree. Do you agree that a spherical black body object, in a vacuum, will radiate energy in all directions in the form of photons, according to the fourth power of its temperature (i.e. the Stefan-Boltzman law), with the spectrum of radiation governed by Plank's law? If you answer my questions directly, I am sure we will soon reach agreement.

Claes: Yes, if you by a vacuum mean a surrounding environment at 0 K. Is that so? And if you by "photons" mean electromagnetic waves. Is that so?

Gavin: By vacuum, I did indeed mean that the surrounding environment is at 0K. By "photon" I was referring to "an elementary particle, the quantum of light and all other forms of electromagnetic radiation". Do you still agree, given those clarifications?

Claes: Yes, go on.

Gavin: Thank you. A second black-body object (lets call it "B" and the first "A") is then introduced. For convenience make B spherical, with the same radius as A, and placed a short distance from, but not touching A. Would you agree that B also radiates photons in all directions according to the fourth power of its temperature (Stefan Boltzmann) with a spectrum given by Plank's law?

Claes: No, since with A present, the environment of B is not a vacuum at 0 K. And then?

Gavin: The intensity and spectrum of black body radiation depends only on its temperature, as given by the Stefan-Boltzmann and Plank laws, which is why black-body objects are a useful idealization. Can you provide a reference to a derivation for black-body radiation that explicitly states the dependence on environmental temperature?

The thought experiment we are conducting doesn't depend on this point, but we appear to have identified a point of divergence, so it would be useful to understand the source of the disagreement.


Claes: It is certainly very natural to expect that the state of the environment is of importance. We just agreed that a body emits according to Planck's law into a vacuum at 0 K. Don't you remember that? If you claim that the environment is of no importance, then you have to back that with strong evidence, since it is such a strange utterly surprising statement. So what is your evidence? Planck's proof of his law only counts degrees of freedom in a cavity and says nothing about independence of surrounding environment, and thus cannot be used as positive evidence of your claim about independence. Right?

Gavin:Regarding the equivalence between cavity radiation and black body radiation, there is a nice explanation here by Prof. Alan Guth of MIT (it is from an excellent course on the early universe that is well worth watching). Essentially the radiation within a cavity is described by Plank's law, but if you were to put a black body into that cavity and wait for it to reach thermal equilibrium, then it must radiate according to Plank's law as well in order for the incoming energy from the cavity radiation to match the outbound black-body radiation, and like cavity radiation, only depends on temperature. Note in this case, the derivation definitely doesn't depend on the environment being at 0K. Prof. Guth also specifically states that its radiation wouldn't change if you took it out of the cavity (at least until it cooled, but then its radiation would be according to SB and P laws at the lower temperature). Prof. Guth is a leading expert in cosmology, where black-body radiation is an important concept (e.g. cosmic background radiation), so I suspect his understanding of this topic is reliable. However, if you have a reference to a derivation of the Plank and Stefan-Boltzmann laws that detail the sensitivity to the environment, then I would happily read them. Can you supply me with such a reference?

Now the environment is certainly important in determining the warming or cooling of the black body objects, which is why I made the simplifying assumption of a vacuum at 0K. However, whether the bodies warm or cool does not depend solely on their intrinsic radiation, as we shall see later in the thought experiment, so I see nothing unnatural about the intensity and spectrum of radiation depending only on temperature.

Let us assume that object B is a little cooler than object A, would you agree that it (B) radiates photons equally in all directions (being a spherical black body object)?

Claes:The version of SB you find in engineering texts, which is relevant to atmospheric radiation, states that the heat transfer between a body A at temperature T_A and a body B at temp T_B is given by Q = sigma (T_A^4 - T_B^4) with A warmer than B and transfer of Q from A to B. The dependence of the environment is here obvious, right? Do you accept this version of Stefan-Boltzmann?

Gavin:That equation is for heat transfer, not for the radiation of photons from a black-body object, we will get on to transfer later. For the moment, I am trying to establish your position on the intensity and spectrum of photons radiated from B as that is important in explaining the transfer.

Let us assume that object B is a little cooler than object A, would you agree that it (B) radiates photons equally in all directions (being a spherical black body object)?


Claes:No, as I have said, the presence of A will influence the radiation from B, since A is part of the environment of B. And yes, we are speaking about heat transfer by radiation, and nothing else, right? And it is better to leave out cosmology, inflation, Big Bang and multiverse, since it is is of little importance concerning atmospheric radiation, right?

Gavin: As I have asked ClaesJohnson twice for a reference giving derivations of the Plank and Stefan-Boltzman law that detail the sensitivity to the environment, and none has been provided, I will have to leave that issue to one side for the moment.

Claes, my previous question referred only to the direction of photons emitted by object B, do you agree, that being a spherical black body object, B will emit photons equally in all directions?

BTW, I should have said more explicitly that I do agree with the formula "Q = sigma (T_A^4 - T_B^4)" for radiative transfer; I certainly do.

Claes: If we do agree about SB as stated, then we are on speaking terms. The presence and dependence of the environment is clear in this formula, right? In particular, we have our previous agreement in the special case with T_B = 0 K, right? What I have asked you about is reference to a statement of independence of environment. What is your evidence? We have agreed on dependence and now it is up to you to deliver contradictory evidence of independence. What is it?

Gavin: "The presence and dependence of the environment is clear in this formula, right?" no, as I said, that equation is about radiative transfer between objects, not the radiation from the objects themselves.

"What I have asked you about is reference to a statement of independence of environment. " I have already provided two, the second being the lecture by Prof. Guth, whom most would regard as being well qualified on the subject.

"We have agreed on dependence and now it is up to you to deliver contradictory evidence of independence" I have already clearly stated that the environment is relevant to transfer, but not to the intrinsic radiation of the object (indeed the transfer equation can be derived from intrinsic radiation being independent of environment).

Now this is the fourth time I have asked this question without a direct answer, I repeat:

Claes, my previous question referred only to the direction of photons emitted by object B, do you agree, that being a spherical black body object, B will emit photons equally in all directions?


Claes: Again, are we discussing heat transfer by radiation, or something else? What, if so?

Gavin: ClaesJohnson we are discussing the radiation of photons from a black body object, with the intention of explaining the nature of radiative transfer from one black body to another in due course.

Claes, my previous question referred only to the direction of photons emitted by object B, do you agree, that being a spherical black body object, B will emit photons equally in all directions?

Claes: No, the environment of B will influence the heat energy radiated by B.

Gavin: ClaesJohnson, O.K. so does B emit any photons that strike (and are absorbed) by A?

Claes: What does that have to do with the heat transfer by radiation between A which we are discussing? Or are you discussing something else? If so, what?

Gavin: In order to explain my argument to you, I need to properly understand your objection. The best way to do this is to ask questions that allow you to unambiguously state your position in a way that I will understand. You may not understand the relevance of these questions, but I suspect it will be clear to most readers with a background in physics, but the fastest way to reach agreement is simply to give a concise and direct answer to the question. So, please give a direct answer to the question: does B emit any photons that strike (and are absorbed) by A?

Claes: I cannot answer because I do not understand the physics of "photons that strike and are absorbed by A". Again, are we discussing heat transfer by electromagnetic waves? If not, what is it you are discussing?

Gavin: A black body is an idealized physical body that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence (source).

Radiation and absorption of photons is the basic mechanism by which radiative transfer occurs. Absorption simply means that the photon no longer exists and the energy that it carried has been transferred to the body that absorbed it. So does B emit any photons that strike (and are absorbed) by A?

Claes: Gavin, if we agree about SB as stated in engineering literature, why is this not enough to describe atmospheric radiative heat transfer? What more do you want? My analysis of blackbody radiation is exposed at https://computationalblackbody.wordpress.com/

Gavin: Claes, as I have pointed out to you before, that equation is for radiative transfer. In order to answer your question about the second law of thermodynamics and the back-radiation, you need to understand how that transfer arises from an exchange of energy in either direction. It is a shame that you have been so unwilling to give direct answers to straightforward questions, so I will explain it to you.

So, the conventional interpretation of thermodynamics would indicate that the warmer of the two bodies would radiate energy (in the form of photons) at a rate given by the Stefan-Boltzmann law, i.e.

Ja = sigma*Ta^4,

where Ja is the power radiated from A, Ta is the temperature of A in Kelvin and sigma is the Stefan-Boltzmann constant. Now, A being a spherical black-body will radiate photons evenly in every direction, however a proportion of these, which we will call c, will be traveling in the right direction to intersect with B, which being a black body will absorb them. The rate at which energy is received at B due to this flow of photons from A is

c*Ja = c*sigma*Ta^4

Similarly, B will radiate energy at a rate given by the Stefan-Boltzman law, such that

Jb = sigma*Tb^4,

where Jb is the power of the radiation from B and Tb is the temperature of B. Now by symmetry (as I have made the two objects spheres of identical radius), the proportion of the radiated photons from B that intersect with A is the same as the proportion of photons emitted by A that intersect with B, i.e. c. Thus the rate at which energy is received at A due to this flow of photons from B is

c*Jb = c*sigma*Tb^4

Now let's consider the gain of energy by the cooler body, B. It has gained energy from A at a rate c*sigma*Ta^4, but has lost energy to A at a rate c*sigma*Tb^4. The transfer of heat between the two, is just the difference of these quantities, i.e.

Q = c*sigma*Ta^4 - c*sigma*Tb^4

or in other words

Q = c*sigma*(Ta^4 - Tb^4)

which is the usual "engineering" representation of the Stephan-Boltzmann law of radiative transfer. The important thing to note is that this arises perfectly naturally from the intensity of black-body radiation depending solely on its absolute temperature. There is no need for some unspecified physical mechanism that influences the direction in which a black body radiates; it just radiates in all directions, and the net flow of energy conforms precisely to the second law of thermodynamics. This is what the footnote in Clausius' book describes, and the basic idea has been well understood for a long time.

So, if a spherical black body does not radiate photons equally in every direction, but is affected by its environment, please explain the physical mechanism by which this is achieved.

Footnote, I am assuming that the constant c has been aggregated with the Stefan-Boltzmann constant in ClaesJohnsons' equation. The constant c depends on the size of the objects, their shape and how much of the radiation from one object is able to fall on another. I have made the scenario in the thought experiment symmetrical so it is easy to see that c is the same for both bodies, although this is true without the exact symmetry.

Claes: I have asked you if an engineering version of SB (or Planck) is enough to describe atmospheric radiation, and if not, what is missing. If you agree that it is basically enough, then we have a common standpoint and we can go on to specific questions concerning the physics of the so called "greenhouse effect". If you insist that it is not enough, then I want to see your arguments supporting this view.

There are endless questions that can take a lifetime or more to answer, such as: What is an "infrared photon"? What physical laws does it follow? How does it travel through space? In straight lines? What is the process of "absorption/emission of an "infrared photon". In which direction is it emitted? Is an "infrared photon" particle or wave? What is its lifetime? How does it interact with other "infrared photons". Are "infrared photons" like bullets traveling through space? If so, what happens when two "infrared photons" meet? Is heat transfer between two bodies carried by two opposite streams of "infrared photons" back and forth between the bodies? If so, what is the mechanism that guarantees that the net heat transfer is from warm to cold? What is the wave function of the multiverse? Et cet, et cet...

But all these questions are irrelevant as concerns the physics of the "greenhouse effect", if the engineering version of SB/Planck (which we have agreed is valid) is enough to describe atmospheric radiative heat transfer.

So I ask you again if we can take this version as a common ground and then proceed to the real questions of importance concerning the physics of the "greenhouse effect", with a basic question being climate sensitivity as the amount of global warming from doubled CO2.

Is this OK to you? Or do you insist that questions of the type I have listed above, have to be answered (by me) before we can can come to the point? And in particular before you will give an answer to my original question about the statement in the video of week 3 that "although heat moves in all directions.." (Is this statement connected to an idea of opposite streams of photons between bodies?)

I expect to get a clear answer to my clearly stated questions, and not just more questions from you to me, which I cannot answer (and probably nobody else).

To meet a question by a battery of counter-questions is a way to avoid answering the original question. I am sure you would not like to resort to this form of discussion trickery, right?

I also ask you if you have looked at the web site on Computational Blackbody Radiation I referred to and if you have read and understood the arguments there presented, and if you have some comments or questions concerning the material?

In short: If net heat transfer from warm-to-cold is what matters, why insist on net heat transfer as the difference of two opposite heat transfers warm-to-cold and cold-to-warm, where the latter appears to violate the 2nd law?

I have given my argument for net transfer as a property of stability (connected to the 2nd law). Transfer as the difference of two opposite gross transfers is an unstable process, since small differences in gross transfers can shift the sign of the net transfer, and thus violate the 2nd law. The only way the 2nd law can be upheld with transfer as difference of opposite gross transfers, is that the opposite transfers somehow are linked, but that contradicts your idea that the opposite transfers are independent of each other. Do you see this?

Gavin: ClaesJohnson wrote

"I have asked you if an engineering version of SB (or Planck) is enough to describe atmospheric radiation, and if not, what is missing. If you agree that it is basically enough, then we have a common standpoint and we can go on to specific questions concerning the physics of the so called "greenhouse effect"."

I have already explained why more is required, when I wrote in the previous message:

"Claes, as I have pointed out to you before, that equation is for radiative transfer. In order to answer your question about the second law of thermodynamics and the back-radiation, you need to understand how that transfer arises from an exchange of energy in either direction."

When we reach agreement on that point, then we will have a common standpoint to discuss the greenhouse effect (and specifically why there is no violation of the second law of thermodynamics).

So, do you accept that the engineering version of SB can be derived (as shown in my previous comment) as the net result of a bi-directional transfer of energy from A to B and from B to A, where the radiation from A is determined solely by is absolute temperature Ta (according to SB), and the radiation from B determined solely by its absolute temperature Tb (according to SB)?

To meet a question by a battery of counter-questions is a way to avoid answering the original question. I am sure you would not like to resort to this form of discussion trickery, right?

Please lets leave rhetoric out of this discussion. I have asked one question in this comment, and one only, please give a direct answer.

Claes: What is your question to me again? I will certainly try to answer if I only understand what you ask. Your answer to my question is that the engineering version of SB/Planck is not enough to (mathematically) model atmospheric radiative heat transfer. But you did not answer my follow-up question to your answer, namely, what more you then need to (mathematically) model atmospheric radiative heat transfer? What additional physical law do you need for this purpose?

You did not either answer if you have looked at the web site I gave. Have you? If so any reaction? And what about my original question?

You say that there are certain things I need to understand in order for you to answer my questions. I don't see that my understanding, whatever it means, is necessary in order for you to answer my questions? Would it not be possible for you to simply answer my clearly stated questions, regardless of my state of mind?

Is it necessary for me to give a complete account of my inner status and thoughts in order for you to answer my questions in my role as student in a course that you are giving on edX?

Isn't the role of a teacher to answer questions from students concerning the material presented by the teacher, rather than subjecting the students to interrogation to see if they carry ideas which the teacher does not like?

And again, what more than the engineering version of SB/Planck do you need to model atmospheric radiative heat transfer??? My view on this question is presented as Unphysical Schwarzschild vs Physical Model for Radiative Transfer at http://claesjohnson.blogspot.se/2015/04/unphysical-schwarzschild-vs-physical.html


Gavin: ClaesJohnson, you seem to have asked multiple questions in your comment, I am happy to answer them one at a time. Please select the technical/scientific question you would like me to answer.

In an earlier comment, I derived the engineering form of the SB equation for heat transfer as being the net result of a bi-directional transfer of energy due to the radiation from each body. This is not a new idea, in his book "The Theory of Heat Radiation", Max Plank (c.f. Plank's law) states:


A body A at 100C emits toward a body B at 0C exactly the same amount of radiation as toward an equally large and similarly situated body B' at 1000C. The fact that the body A is cooled by B and heated by B' is due entirely to the fact that B is a weaker, B' a stronger emitter than A.

This makes it very clear that Plank's conception of heat transfer is of a bidirectional transfer of radiation, both from warmer to cooler and from cooler to warmer, with the transfer of heat depending on the net difference in the two flows. Note Plank also specifically states that the radiation from A is not dependent on the temperature of the body on which the radiation will fall.

So to repeat my question:
...do you accept that the engineering version of SB can be derived (as shown in my previous comment) as the net result of a bi-directional transfer of energy from A to B and from B to A, where the radiation from A is determined solely by is absolute temperature Ta (according to SB), and the radiation from B determined solely by its absolute temperature Tb (according to SB)?

Claes: Of course the one-directional SB can be derived from a two-directional version by trivially taking the difference. But that does not say that the two-directional is correct, right?. The one-directional version may be the correct physical law, while the two-directional may still be non-physical. Confirming an assumption by observing a consequence is one of the logical fallacies, right?

So I have answered your question, and now to my question: What more than the engineering version of SB/Planck do you need to mathematically model atmospheric radiative heat transfer? What additional physical do you need for that purpose? The question is clearly stated and I expect a clear answer.


Gavin: Claes, again you have asked multiple questions (note there are three question marks in your comment). However I will address them in turn, on this occasion (I assume the second was rhetorical):

"Of course the one-directional SB can be derived from a two-directional version by trivially taking the difference. But that does not say that the two-directional is correct, right?."

No, no in itself. However the quote I gave from Plank clearly shows that Plank's conception of heat transfer was the net result of a bi-directional flow of energy. Similarly Clausius' book clearly indicates that a bi-directional flow of heat is completely consistent with the second law of thermodynamics, provided the net flow is from hot to cold. More importantly, I would argue that there is no plausible physical mechanism that can explain how a body can modify its radiation to avoid its radiation being absorbed by a warmer body. My derivation requires no such assumption as the radiation of a body depends only on its local state (specifically its absolute temperature). So my question for this message is: "What physical mechanism allows a body to alter its radiation to avoid emitting photons that reach a (possibly moving) warmer body?"

"What more than the engineering version of SB/Planck do you need to mathematically model atmospheric radiative heat transfer?"

I have answered this question several times already. The net transfer of heat is depends on the difference in the energy exchanged between two bodies (in this case the atmosphere and the surface). Thus it is necessary to consider the amounts of energy radiated by each component and absorbed by each component separately. The greenhouse effect does not violate the second law of thermodynamics because the energy transferred by back radiation from the atmosphere to the surface is "compensated" (as Clausius, in translation, would say) by a larger transfer of energy in the other direction, in the form of IR radiation from the surface.

The key point is that if you accept that a bi-directional exchange of radiation doesn't violate the second law of thermodynamics, provided the net flow is from hot to cold, then the greenhouse effect doesn't violate it either. If you do not accept this, then you need to show that the engineering form of the SB law is inconsistent with the interpretation as the net result of a bi-directional exchange of radiation. However you appear already to have conceded this point

"Of course the one-directional SB can be derived from a two-directional version by trivially taking the difference".

Claes: I asked you: What additional physical law, in addition to the one-directional engineering SB/Planck law we have agreed on, do you need to mathematically model atmospheric radiative heat transfer? What is your answer?

Gavin: Claes, I have already answered that question. We must use the physical laws governing the radiation of black-body objects (at least to begin with), which is the Stefan-Boltzmann law for radiation, i.e. j* = sigma*T^4. From this we can straight-forwardsly derive the "one-directional engineering SB/Planck law" as the net result of an exchange of radiation between two bodies. Under this interpretation of the "engineering SB law", the greenhouse effect does not violate the second law of thermodynamics as the radiation from the warmer surface to the cooler atmosphere is greater than from the atmosphere to the surface. Therefore the "one-directional" net heat transfer of the "engineering" SB law is from warmer to cooler, as required by the second law of thermodynamics.
This question ought to have a yes or no answer, and will help me to understand your position if you give an unequivocal answer. Plank writes:
A body A at 100C emits toward a body B at 0C exactly the same amount of radiation as toward an equally large and similarly situated body B' at 1000C. The fact that the body A is cooled by B and heated by B' is due entirely to the fact that B is a weaker, B' a stronger emitter than A.
My one question for this comment is: Is any of the radiation emitted by B (at 0C) absorbed by A (at 100C), "yes" or "no"?

Claes: This statement of Planck lacks physical reality. Nature does not play with opposite equally large quantities, which are independent, yet always keeping one bigger than the other to not violate the 2nd law. You cannot accept anything that Planck says without yourself judging if that is correct or not? Science is not parrot science where you simply repeat what is written in book or stated by some since long dead scientist.

Again: what additional law is required to mathematically model atmospheric radiative heat transfer, beyond the one-directional SB/Planck law we have agreed on? Is it two-way heat transfer? If so, what equation does that effectively bring into the mathematical model? Have you read my post about Schwarzschild's (unphysical) equations based on two-way transfer? If not, do that and give your view on the necessity of Schwarzschild's model. OK?


Gavin: Claes, I have already answered your question repeatedly. The additional physical law that is required is the Stefan-Boltzmann law of radiation: j* = sigma*T^4. For the reasons, see my previous answers.

Now please give a direct answer to my previous question, I repeat:

Plank writes:A body A at 100C emits toward a body B at 0C exactly the same amount of radiation as toward an equally large and similarly situated body B' at 1000C. The fact that the body A is cooled by B and heated by B' is due entirely to the fact that B is a weaker, B' a stronger emitter than A.

My one question for this comment is: Is any of the radiation emitted by B (at 0C) absorbed by A (at 100C), "yes" or "no"?

You earlier wrote:

To meet a question by a battery of counter-questions is a way to avoid answering the original question. I am sure you would not like to resort to this form of discussion trickery, right?

Note that in my previous comment I asked precisely one question (only one question mark), but in your reply, you did not give an answer, but you asked multiple questions (I count six question marks!).

Claes: My answer is no.

Gavin: Thank you, that is interesting. Consider a third body B'' at 50C, which is of a similar size to B and again similarly situated. I am assuming that since B'' is also cooler than A, you would say that no radiation from B'' is absorbed by A either. Feel free to correct me if this assumption is incorrect. My question is, does A emit a different amount of radiation towards B than it emits towards B''?

Claes: Radiative heat transfer between bodies is described by the one-directional SB law we have agreed is valid.

Gavin: That is not a direct answer to the question, A either does emit a different amount of energy towards B than it emits towards B'', or it does not. Which is it?

Claes: SB gives the answer to your question. This is an exercise you can do yourself. After all you are the teacher and should know.

Gavin: The reason that I am asking for an unequivocal answer is that I intend to demonstrate a contradiction and do not want to leave room for equivocation after it has been established. If you are confident of your position, you ought to be eager to state your position in completely unequivocal terms. So I ask again:

"does A emit a different amount of radiation towards B than it emits towards B''?"

"yes" or "no".

Claes: A as warmer transfers heat to B and B" according to SB. If B" is warmer than B, then less heat energy transfers from A to B" than to B.

Claes: You claim that in addition to SB in the form Q = sigma (T_A^4 - T_B^4) with T_A > T_B, you need an SB of the form Q = sigma T_A^4. But the latter is included in the former if you set T_B = 0. So why is the extra SB needed?

Claes: Gavin: While you are thinking, I hope you also remember to answer my original question about the meaning of the statement "although heat moves in all directions..." in a video of week 3.
Gavin: I have now asked ClaesJohnson a straight-forward "yes"/"no" question three times, and each time have recieved an evasive response. As I have already explained why an indirect answer would be indicative of evasion, I think it is reasonable to conclude that the evasion was deliberate.

ClaesJohnson subsequently attempted to divert the discussion away from a line of inquiry that will demonstrate a contradiction in his position by repeating a question from earlier in the discussion that has already been answered (repeatedly). Again this is evasion.

Socratic method (a form of enquiry based on asking and answering questions) is an excellent means of resolving scientific disagreements, provided that both parties engage in the exercise in good faith. This cross-examination allows misunderstandings to be resolved and exposes the weaknesses in either argument. Evading direct questions is a clear indication that someone is unwilling to change their views, regardless of the evidence or opposing arguments presented. In this case, there is little point in continuing the discussion, and the observers can draw their own conclusion from the evasion.

Ulimately, if ClaesJohnson refuses to look at anything other than the "one-direction" SB law for radiative heat transfer, and is unable to understand that this arises as the net result of a bi-directional transfer of energy (as illustrated by Planck's example), he will be unable to understand why the greenhouse effect doesn't violate the second law of thermodynamics. However, he can't say that this has not been explained to him.

Claes: Gavin, I think we have come to the end of our discussion. Yes, it is true that only the uni-directional SB makes sense to me and to physics. You have not been able to give any scientific support to a "greenhouse effect" based on two-directional heat transfer including "back radiation" with heat transfer from cold to warm, and neither have you been able to explain the obvious violation of the 2nd law in such a process. This means that a "greenhouse effect" based on "back radiation" is nonphysical illusion. The result is that the course lacks sufficient scientific basis and should be closed.

Gavin: ClaesJohnson If only the uni-directional SB makes sense to you, then perhaps you should not obstruct attempts to explain the bi-directional energy transfer required for an understanding of back radiation by the sort of evasive behaviour you have demonstrated during this discussion.
Claes: I have declared my standpoint very clearly. The evasiveness is yours. I am surprised that edX offers a platform for the kind of propagandistic disinformation the course presents. In any case the "denial" will not be affected by the course.

27 kommentarer:

  1. Claes I tend to think that your comment is too long and does not come to the point.
    The proposition that a body at some temperature in a vacuum at 0 degree K radiates in all direct is nonsense because it does not exist. Nothing exists at 0K. Space outside the solar system is said to be at 4K but it is not a complete vacuum as there are other bodies, gases, light waves, electrical waves, radio waves and magnetic fields present which carry energy. In our solar system the sun radiates energy but it is not equal in all directions all the time. Photographs and measurements show that there are cooler temperature sun spots and higher temperature solar flares. The second point of an additional body B which is lower temperature than the first body A is also not physical. The sun does not receive heat from the earth. When the moon comes between the earth and the sun in an eclipse the cold side of the moon does not radiate heat towards the earth. Postulate 4 of Thermodynamics (or the 2nd Law of Thermodynamics) clearly states that the entropy change of a system and its surrounds is positive (ie heat transfer occurs only from high temperature to lower temperature)

    SvaraRadera
    Svar
    1. Neal J. King24 maj 2015 04:21

      - I agree that the restriction to 0 Kelvin doesn't make sense.
      - The exterior surface of the Sun is not a blackbody precisely because the temperature varies over the surface; but for relatively short distances in the interior, the radiation distribution is blackbody because of the great opacity of the solar stuff.
      -The 2LoT is perfectly satisfied when the net power transfer is from hotter to colder. If you think it violates the 2LoT to have power flowing from colder to hotter - but being over-powered by the hotter-to-colder flow - then produce a perpetuum mobile of the 2nd kind. That is the criterion for a 2LoT violation.

      Radera
  2. We agree, but I think my question if the one-directional version of SB is enough to model atmospheric radiative heat transfer, is a good one, and after a long intro we have now come to this central point. See also the post on the unphysical Schwarzschild equation based on the two-directional version.

    SvaraRadera
  3. Claes: I think you should give an honest answer to Mr Gavins question: "Is any of the radiation emitted by B (at 0C) absorbed by A (at 100C), "yes" or "no"?" This is a qey question. I have earlier proved to you that a blackbody can´t distinguish between radiation coming from a colder or warmer body than itself, if the frequency is the same. And with overlapping frequency spectra, the frequency can be the same!

    SvaraRadera
  4. Your own body can well distinguish radiation from a cold or warm wall in the room you are sitting in.

    SvaraRadera
  5. "Your own body can well distinguish radiation from a cold or warm wall in the room you are sitting in." I agree, but that describes the net transfer. A single atom, ion or molecule can not decide if radiation with frequency f comes from an atom (ion or molecule) in a warmer or colder body. If frequency f lies within its own spectrum it can absorb that radiation whether it comes from a warmer or colder body.

    SvaraRadera
  6. From where do you have this information oncerning the perception and decision of an atom?

    SvaraRadera
  7. "From where do you have this information oncerning the perception and decision of an atom?"
    From Kirchoff´s law of thermal radiation, and this is (should be) wellknown stuff for you.

    SvaraRadera
  8. Please state this law of Kirchhoff and how it describes what atoms "decide" to do.

    SvaraRadera
  9. The atoms, ions or molecules in a blackbody that emit radiation at frequency f can also absorb radiation at frequency f.

    SvaraRadera
  10. Under your concept that the hotter objects radiate towards the cooler objects but not the reverse, the radiation field gets too "spikey", in a way incompatible with Maxwell's equations.

    SvaraRadera
  11. you are mixing up cards claes: you are the one claiming that an atom has a perception of the origin of radiation and takes a decision, not Lasse H. So you should give a justification of that.

    SvaraRadera
  12. Actually, I find it very hard to understand how the emission could be affected by the temperature of the receiver.It looks like spooky action-at-a-distance to me. Whereas if each body is emitting in all directions as a function of its own temperature only, it is obvious how the net transfer depends on both temperatures.

    SvaraRadera
  13. My view is that heat transfer is a resonance phenomenon of two bodies in contact by electromagnetic waves, where heating of one of the bodies occurs if and only if the other body is warmer and thereby can resonate at a frequency above cut-off for the first body. The analysis behind this view is presented on the web site Computational Blackbody Radiation. And yes, in a resonance ohenomenon between two bodies they have contact. It is the same phenomenon as two tuning forks in resonance, which may look as "spooky action at distance" but is not spooky at all, just resonance mediated by acoustic waves. I suggest that you consult the web site before continuing speculations about what atoms can decide or know.

    SvaraRadera
    Svar
    1. Neal J. King24 maj 2015 04:07

      It seems a lot more complicated than what old Planck was thinking: That every blackbody just spews out radiation with an intensity and frequency distribution characteristic of its temperature.Then the net flow is from the hotter to the colder, because the hotter is spewing out more. Seems pretty basic to me.

      Also, it is famously said that "In space, no one can hear you scream": No air, no acoustic waves. But somehow EM radiation manages to conform to the 2LoT.

      Radera
    2. Neal J. King24 maj 2015 04:32

      Honestly, Claes, you haven't said anything interesting enough to make it worth the effort. Your suggestions are more complicated and less plausible than what normal physics says. At least with that EM-Drive nonsense a year ago, or the electrical engineer who thought he had discovered a violation of conservation of angular momentum in relativity, there was an apparent (though mistaken) conundrum. You just appear to be thinking about it the wrong way. That is all.

      Radera
    3. If you prefer simplistic fiction instead of real physics, then there is nothing I can do.

      Radera
    4. Claes:
      I have read the real physics as taught by Planck on heat radiation, and generations of physicists on quantum mechanics. If you want to convince people to spend time critiquing your version of events, you need to convince them that there is some point. What actual insight or convenience does your approach offer? What conceptual clarification or range of data are brought in? If you have nothing to offer, nobody will take any interest.

      Radera
    5. The main idea is to replace statistics of particles, which is not physics, by physics as analog computation with finite precision, which may be physics. I know that physicists have been brainwashed by particle statistics and it is very difficult to get across a different idea. I can only hope that some people are willing to reconsider what they have been taught. Maybe you are one of them.

      Radera
  14. in two tuning forks in resonance there is no temperature dependence of the forks. just the frequency is of interest.
    you are here claiming that the resonance has a cut off that depends on the temperature. i suggest that you try to explain how this is physically possible in a way that at least one other person in the world can understand, instead of referring everybody to read your very long work.
    the situation proposed by Gavin is very simple (and has been proposed to you many times before, always with the same tactic of not answering from your side).
    how does it work for the overlapping frequencies? how can the body decide which are to be absorbed and which are not?

    SvaraRadera
  15. Repeating a tune from the radio on a piano is increasingly difficult as the temperature of your fingers decrease until cut-off where the fingers are so stiff that no keys can be pressed and the repetition/reradiation comes to nothing. But during the struggle the fingers will heat up and eventually the tune can be repeated/reradiated and the heating stops.

    SvaraRadera
  16. and what has this to do with what we are talking about?
    there is no resonance involved, there are no waves involved.
    in the analogy with the forks you disregard the difference between mechanical waves and electromagnetic waves for the sake of making a point. now you even compare with the the reproduction of a tune from a radio by a person.
    how about a serious, scientific answer?

    SvaraRadera
  17. Red herring...

    SvaraRadera
  18. If you do not care to read my scientific work and is not willing to use your fantasy to follow an everyday analogy, I cannot help you, and you should pose your questions to someone else instead.

    SvaraRadera
  19. that is not an everyday analogy, for the simple fact that there is no analogy to the case at hand and that it very rarely, let be everyday, happens that somebody who is trying to reproduce a tune on piano from a radio stops doing so because their fingers are paralysed from cold.

    SvaraRadera
  20. Claes,

    which law of physics prevents a specific photon emitted by a colder body being absorbed at the hotter body. It is not the 2LoT. The 2LoT is a law that summarizes our experience.
    Heat transfer is an energy transport that increases the number of accessible states in the system that receives the energy and diminishes the number of accessible states in the system that looses the energy. This means in turn that photons being transferred from cold to hot do not violate the 2LoT, because it is the net energy transfer that increases the energy and the entropy and therefore the number of accessible states in the colder body and the net loss of energy in the hotter body that decreases the energy and the entropy and therefore the number of accessible states.
    This microscopic picture is totally able to explain the observed phenomena.
    Your picture however as I understand it requires the colder body to have the "knowledge" that there is a warmer body in order to stop emitting photons. This "knowledge" is unphysical even so you claim a resonance effect.

    SvaraRadera
  21. See later post for my answer to your science fiction about photons and micro states.

    SvaraRadera