söndag 6 februari 2011

Black Magic of Stefan-Boltzmann's Law for Two BlackBodies


gives the spectral radiance (flux of radiant energy) from an ideal blackbody as function of temperature T and frequency nu. Integrating over frequency gives the total radiance R as Stefan-Boltzmann's Law

(1) R = sigma T^4

with sigma a certain constant referred to as Stefan-Boltzmann's constant.

In the engineering literature the following Stefan-Boltzmann law is presented for the radiant energy exchanged between two blackbodies 1 and 2 of temperature T1 and T2:

(2) R1 = sigma T2^4 - sigma T1^4, R2= - R1 = sigma T1^4 - sigma T2^4,

where R1 is the net radiance received by body 1 and R2 that received by R2.

How is the formula (2) derived? From a Planck Law for two interacting blackbodies? No, it appears to be an ad hoc law derived by the following simple ad hoc argument:
  • Stefan-Boltzmann's Law (1) for one blackbody gives the radiant energy from a blackbody at a certain temperature T into a receiving background at 0 K.
  • Now, if we have two blackbodies, both will radiate into the same background at 0 K as if the other body was not present.
  • To get the effect of the two bodies, just sum the contributions and you get (2).
  • Voila!
But this is an ad hoc argument without specified physical realization of the summation, thus a black magic argument. This is the argument behind the "backradiation" from a cold atmosphere to a warm Earth surface underlying CO2 climate alarmism. Black magic.

In my new derivation of Planck's Law in Slaying the Sky Dragon, the temperature of the background enters and the radiant energy exchanged between two blackbodies is derived from basic physical principles and not from ad hoc summation without physics.

PS I don't say that (2), as a formula for the net radiative heat exchange, is wrong. What I seek is a derivation of the formula which can help understanding of the physics involved, without resorting to statistics which at least to me is not understandable.

49 kommentarer:

  1. hmm.. even i have always had a doubt about using Stefan-Boltzmann's law like in the conventional books.. Think, now I should read up your book seriously!.

    SvaraRadera
  2. The next step should be to measure spectral radiance with a suitable detector.

    This might not be as easy as it seems since most detectors simply assume the above equation and calibrate using the SB law.
    This produces a circular argument.

    What is required is an instrument which is independent of the Stefan-Boltzmann's Law

    Bryan

    SvaraRadera
  3. Yes it is circular, as is "backradiation". An instrument measuring net radiative transfer shoud be easy to make: Just put in the radiation at hand and measure how much it heats up, like putting a pot of water in the sun shine, or in the glow of the night...

    SvaraRadera
  4. Equation (2) looks perfectly natural. Each body emits electromagnetic waves into vacuum and then at some point they meet the other object.

    SvaraRadera
  5. Claes wrote: "In my new derivation of Planck's Law in Slaying the Sky Dragon, the temperature of the background enters and the radiant energy exchanged between two blackbodies is derived from basic physical principles".
    Then you also have prooved that the earth will radiate less (getting warmer) if some gases around the earth radiate IR (equal to having an equivalent T> 0´K). I have read somewhere that this increase in earth temp is 33´K. Right? And these gases are mostly water vapour and CO2.

    SvaraRadera
  6. Claes, Brilliant analysis as usual. We need this in the next Slayers book!

    SvaraRadera
  7. Just because something "looks perfectly natural" does it mean that it is real physics. Eq (2) should be derived from basic principles, unless it is just to be viewed as an ad hoc rule of thumb, or black magic. Right?

    SvaraRadera
  8. The Earth will not radiate less, but is true that water vapor and gravitation and other things effect global climate. The effect of CO2 appears to be nil.

    SvaraRadera
  9. Basic principles: (i) net = input - output (ii) superposition of EM waves.

    SvaraRadera
  10. Is that all? Is this supposed to describe the interaction between two systems? As you write it, it is an ad hoc triviality obtained from SB's
    one-body law. What you need is a derivation of the formula from basic principles. This is called theoretical physics, as opposed to black magic.
    But maybe you can derive the formula using statistics?

    SvaraRadera
  11. Yes, that's all. Basic physics.

    If you want to advocate a new magic theory for "two-body Stefan-Boltzmann" which is different from "one-body Stefan-Boltzmann" you first have to demonstrate (by experiment) that 2x"one-body Stefan-Boltzmann" gives incorrect results.

    SvaraRadera
  12. No, I want a theoretical derivation of two-body Planck-SB formula which is based on physical principles and not just ad hoc. That is a derivation based
    on Maxwell's equations.

    SvaraRadera
  13. Claes said:"The Earth will not radiate less...".
    Wrong, at least the water vapour makes the earth radiate less (net radiation)compared with the case where the surrounding is at 0´K. Right?
    Compare with the Wikipedia example in an earlier blog comment where you yourself said that the body decreased its net radiation compared to the case with empty space surrounding.
    And the CO2 effect is not nil, because it has similar properties as the water molecule, but much less than that of water vapour .

    SvaraRadera
  14. Why? What is wrong with the regular Stefan-Boltzmann? Has it been demonstrated that it gives incorrect results?

    SvaraRadera
  15. Well, you have to take thermodynamics into account to say anything because it is a coupled thermydyn-radiation problem. You can argue
    forever about a hypothetical situation with radiation only.

    SvaraRadera
  16. I don't say it is wrong. I just want to derive it from basic principles to understand it and not just passively accept it as some mysterious fact beyond comprehension. Without statistics, because I don't understand statistics. Understand?

    SvaraRadera
  17. What is a hypothetical situation? Two "blackbodies" in vacuum with temperatures T_1 and T_2?

    SvaraRadera
  18. There is no statistics involved in starting from Stefan-Boltzmann and saying that net = input - output.

    SvaraRadera
  19. Is that a derivation without statistics?

    SvaraRadera
  20. Claes said:" ...because it is a coupled thermydyn-radiation problem."
    I agree on that. But to be able to discuss the coupled problem you have to agree on the basics of radiation, and now it seems as we agree on that matter (thanks to your derivation of the twobody SB-law)
    PS Superposition is an exellent tool in radiation physics DS

    SvaraRadera
  21. Starting from SB involves statistics.

    SvaraRadera
  22. What basics of radiation? Not "Backradiation" in any case.

    SvaraRadera
  23. I don't see any statistics: use Stefan-Boltzmann's formula with T_1 and T_2, then subtract.

    Do you agree that you can get the net heat flux between two blackbodies in vacuum from that simple calculation?

    SvaraRadera
  24. Claes said:"What basics of radiation?..."
    I mean the twobody CJ-law, derived from Maxwells eq by pure mathematics (equal to the twobody SB-law), which tells that the earth will get warmer because of less net radiation, depending of absorption of radiation in the surrounding of the earth causing higher temperatures there.
    We don´t need to talk about backradiation, but any materia with T>0´K radiates.

    SvaraRadera
  25. Anders: The formula may be found experimentally to be reasonably correct, but I ask for a derivation of the formula without statistics, from scratch without assuming any SB based on statistics. Isn't this reasonable? At least this is what I seek to do in my blackbody article. Have you read it? Any comment?

    SvaraRadera
  26. Lasse H: If we agree on a SB which only gives net radiative transport (without any "backradiation"), then the next step is to combine it with thermodynamics under gravitation. Right?

    Why do I not want to accept "backradiation"? Because it is unphysical and confuses the understanding of the effect of small perturbations. If you think there is massive two-way energy exchange (of more than 300 W/m2) between the Earth surface and TOA, then you can easily envisage a small 1% perturbation causing "radiative forcing" of 3 W/m2, which is a 5% perturbation of the net radiative transport of 60 W/m2, which is no longer small. You have this way created an alarming big effect out of nothing. If you inflated the two-way exchange to instead 3000 W/m2 (why not) then you could reach a devastating 50% and so on...The comparison with a bubble economy with rotating flow of a fictitious money presents itself and shows the unphysical instability of "backradiation". Right?

    SvaraRadera
  27. No, that is not what you ask for (a derivation of Stefan-Boltzmann)! Your blog post very clearly states that it is wrong to use Stefan-Boltzmann for two blackbodies to compute the net flow of energy.

    So I repeat my question: Do you agree that you can get the net flow of energy between two blackbodies in vacuum from a simple application of Stefan-Boltzmann for T_1 and T2 and then subtract to get the net flow?

    If yes, then I'm happy and you can add a note to your post that you have changed your mind and equation (2) is now correct. If no, please explain why. It is not enough to state that the original derivation of Stefan-Boltzmann itself is wrong and you have a better derivation (which is a separate issue).

    SvaraRadera
  28. The only thing I say is that I derive the heat exchange between interaction two radiating systems (socalled blackbodies) from physical principles such as a wave equation augmented by a third-order time derivative with a small coeff modeling radiation, without resorting to statistics or ad hoc laws. This is what I do, and I have not changed my mind as concerns to this objective.

    I don't say that formulas obtained this way, even if they happen to coincide with some formulas derived by ad hoc or statistics, are wrong. Why should I do that? What I want is to understand the energy exchange and the derivation can help understanding. If you read my article you might open your mind to understanding, or you may find that my argument is mathematically incorrect. Whatever the outcome something productive would have been accomplished, as compared to circular sophistry. So I hope you will read the article and comment. In particular
    it would be most helpful to me to check out if is basically OK, or if I have
    missed something essential. I know of nobody who has read the math of the article and so you would be the first. Why not give it 10 minutes?

    SvaraRadera
  29. I'm not sure I understand your answer. Just to be very clear. Do you say that:

    1. Stefan-Boltzmann is correct and it gives correct results when used to compute the net flux between two blackbodies according to eq. (2)?

    2. It is a mystery that the very simplistic application of 2xStefan-Boltzmann (summation) actually gives the correct result?

    3. You have an alternate derivation which gives the exact same final result C*(T_2^4 - T_1^4)?

    What is your answer(s)? Yes, yes, yes, or something else?

    I'm sure the math in your paper is correct. It usually is, but I'm not sure about the physics (the connection between formulas and reality). If the math is correct and the end result agrees with experiments, then all is well.

    SvaraRadera
  30. Come on: I don't say that Planck's derivation of his formula underlying SB, is necessarily wrong, only that I don't think it is physics, only a "trick" which was also what Planck said. I repeat, what I seek is a theoretical derivation without statistics which can help my understanding and maybe some other people's understanding as well. I have added such a comment on the post., And of course you don't have to read the article, but making comments to a post directly related to the article without having read the article, risk to be irrelevant. Or?

    SvaraRadera
  31. 1 yes.
    2 yes it is a mystery, because SB itself is a mystery (to me, Planck an others) and two mysteries is more mysterious than one unless they cancel. Do they?
    3. The new derivation, if correct, gives many new formulas depending on the specifics of geometry, forcing, et cet. The old formula may show to not describe any complex reality, for examples the interaction between the Earth surface and the atmosphere. But the new approach may do this much better.

    SvaraRadera
  32. It is very easy to get the impression that you think that Planck's derivation is wrong *and* Stefan-Boltzmann's equation is wrong.

    Did you add the PS to your blog post later? It looks like you now agree that Stefan-Boltzmann is correct, but you are very cautious to only state that you don't state that it is wrong. You will not admit that it is, in fact, correct.

    To not state that something is wrong is a very weak statement. For example, I can state that I don't state that your paper is wrong.

    I haven't read the latest version of the blackbody paper. I read the first draft 5 years ago and had some objections then. It would be very interesting to discuss it with you in detail but I'm not sure it is worth the trouble. You are a master of deflecting questions and (slightly) changing the subject (from the correctness of Stefan-Boltzmann to the correctness of your paper) when you suspect that a yes or no answer would get you into trouble.

    SvaraRadera
  33. It is not so meaningful to discuss my article without reading it. Or?
    Why should I have to say that I believe Planks formula is correct if I cannot understand Plancks derivation. As a mathematician I only have reason to believe
    a theorem is correct if I can understand the proof. If I cannot understand the proof, what is my opinion worth? Better then to say nothing. Or is blind faith the codex of math science?

    SvaraRadera
  34. Thanks for you answers! (yes, yes, maybe)

    "It is not so meaningful to discuss my article without reading it. Or?"

    I did not discuss your article. I only responded to your blog post which seemed to make an incorrect statement about Stefan-Boltzmann. Now it seems that we agree on that.

    If you want to have a discussion of your article, maybe you can make blog posts about specific points? It is easier to discuss one detail than an aggregate of details and a sequence of conclusions drawn from all the details. It would also be more interesting if you could demonstrate that your new theory can make predictions that match experiments (with many decimals).

    SvaraRadera
  35. Well to point to perceived incorrectnesses while leaving the context may miss the point. Computations with the new approach is on the way.
    The essence of the analysis is the Fourier analysis covering
    less than half page, so it is pretty focussed, if you are interested
    and can spare 10 minutes. The essence is a phenomenon of
    near resonance where a system of oscillators carries part
    of a forcing, or outgoing radiation, as in the resonating body of
    a guitar or piano. Since you play the piano maybe you have some
    interest to understand what it is all about?

    SvaraRadera
  36. It will be interesting to see what comes out of the computation, in particular if the result agrees with experiments and Stefan-Boltzmann. If it does, I promise that I will read the article.

    As I said, I read an early draft and the most "interesting" point is not the Fourier analysis but the postulation of a new kind of wave equation with a third order time derivative. It seems very ad hoc, but if it predicts correct results (with many decimals, not only in the picture norm) then I will accept it.

    SvaraRadera
  37. Ad hoc? Is the wave equation ad hoc? The third order time derivative was used by Planck in his analysis, if that helps you to accept by authority.
    In fact the model is close that used by Planck as starting point, before he gave in and "in desperation" collapsed to statistics. But you don't have to
    used this sledge hammer and kill both your fingers and physics.

    SvaraRadera
  38. You say that because you do not understand "statistics" it does not belong in physics. What is your explanation for superfluids?
    http://en.wikipedia.org/wiki/Superfluid
    Their behavior is well, and easily, tested in a lab, no fancy instruments are needed, so that is not in dispute. But all working explanations for their behavior are based in quantum physics.

    SvaraRadera
  39. I don't consider quantum physics as necessarily being based on statistics, just
    like Schrodinger.

    SvaraRadera
  40. It isn't true that your math haven't been scrutinized by others yet. This can be seen in the blog-post http://judithcurry.com/2011/01/31/slaying-a-greenhouse-dragon/ where Tomas Milanovic raises some questions about the physical validity of the wave-equation.

    SvaraRadera
  41. Is this what must be read to understand your arguments?
    http://www.nada.kth.se/~cgjoh/ambsblack.pdf
    I think it is a bit tricky to understand.
    Perhaps should I start with the Planck book?
    And also refresh old knowledge in E-M radiation.

    SvaraRadera
  42. @logg Have you ever made any measurements in furnaces and heat exchangers? The S-B equation only applies when modified by an emissivity (see Wikipedia -not the best reference but suffices in this case http://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law)It can be used to estimate heat transfer by radiation when temperature differences are large but
    it is very necessary to use the correct emissivity factors which have been found by measurement and applied by experience. For example- the emissivity of a premixed or nozzle-mixed natural gas flame is only about 0.45 and this is, mainly (over 95%), due to the water vapour. The emissivity can be raised somewhat by design (without premixing) which induces cracking of the methane to carbon.
    When temperature differences between surfaces or a surface and a fluid are low, heat transfer by radiation is less certain and can, in the situation of high fluid Reynolds numbers, be ignored or combined into an overall heat transfer coefficient for forced convection.

    My estimation using engineering formulae and data is that the absorption of radiant heat by the trace CO2 in the atmosphere is insignificant. It should be noted that heat and mass transfer is an engineering subject which clearly many scientists do not understand

    Keep strong
    Cementafriend

    SvaraRadera
  43. The article in the book should be enough.

    SvaraRadera
  44. When you have read "A Conversation with an Infrared Radiation Expert" you will see that his picture of the real world matches what Claes says pretty well.

    I think that when the dust settles, people will understand that Claes is the one who got it right, not the "warmers".

    SvaraRadera
  45. In any way, he is a wise guy, saying this:

    "With so many scientists arguing about the effects of CO2 I am not the one to think I have the answers. I really don’t know what the truth is. And the problem that all these scientists have is that they will never be able to test if their theories are correct, because the time spans are too long. For a theory to be scientifically proven, it has to be stipulated and tested, and the test must be repeatable and give the same results in successive tests for the theory to be proven.
    If not, it is not science, it is guessing.
    More like a horoscope…"

    And I don´t think he supports "coolers" more than "warmers, rather the opposite.

    SvaraRadera
  46. I would remind something about what occurs when two waves traveling in opposite direction interfere between each other, taking into account the simplest case 1-D.
    1- The sum of two counter - propagating waves (of equal amplitude and frequency) creates a standing wave and there isn’t propagation of energy. In other words, using the principle of superposition, the resulting wave may be written as:
    R = Asin(kx+wt) + Asin(kx-wt) = 2Asin(kx)cos(wt)
    that is no longer a travelling wave because the position and time dependence have been separated.
    2 – If the two waves haven’t equal amplitude, then the global effect is:
    R = (A + B)sin(kx+wt) + Asin(kx-wt) = 2Asin(kx)cos(wt) + Bsin(kx+wt)
    that corresponds to a standing wave 2Asin(kx)cos(wt) plus a traveling wave Bsin(kx+wt) which has the propagating sense of (A+B). In other words, the strong wave (A+B) blocks on the way the weak wave (A) and there rests acting only the difference between the two waves.
    That’s, the traveling effect of counter wave Asin(kx-wt) (for us, the back radiation) vanishes on the way and there’s a reduced traveling effect of stronger wave (A+B). Of course, the energy that don’t travels isn’t destroyed but simply turned into potential energy of the stationary wave.
    There is nothing strange. In all phenomena of nature only the strongest survive and it always stifles the weaker, that however isn’t suppressed but only captured because Nature always selects for to operate the most elegant and economic way.
    Michele

    SvaraRadera
  47. Well, fair enough but your model does no contain transfer of energy, only traveling waves, and so is incomplete. Transfer of energy requires absorption and this is what my model contains as a resonance phenomeon
    as a result of near-resonance in a damped wave equation.

    SvaraRadera
  48. I had taken for granted the remainder of the discourse. I now specify better.
    The specific power transferred by a monochromatic EM wave (W/m2)/Hz is given by the Poynting vector, the vector product ExH = E^2 = I(v) (missing the constants). If two waves are counter – propagating and the E vectors is w, the superimposition gives us the EM component
    Ex = E1 + E2cosw, Ey = E2sinw
    Hx = H2sinw, Hy = H1 – H2cosw
    and the magnitude of ExH is (E1 + E2cosw)(H1 – H2cosw) – E2H2(sinw)^2 = E1H1 – E2H2(cosw)^2) – E2H2(sinw)^2 = E1^2 – E2^2 = I1(v) – I2(v).
    In other words, even with a nonlinear relationship, the superimposition can be applied to Poynting vector, i.e. to spectral radiance and to its integral over frequencies.
    There occurs only an energy flux between the two bodies 1 and 2 given by sigma(T1^4 – T2^4) that assumes a specific and rigorous physical significance.
    Thus, the backradiation is truly a non sense.
    Michele

    SvaraRadera
  49. Great writing it is such a good and nice idea thanks for sharing your article .I like your post.
    Thanks.....

    SvaraRadera