måndag 25 juli 2011

Backradiation in Stefan-Boltzmann's Law: Folklore or Science?


Stefan-Boltzmann's Law can be formulated in the following two algebraically equivalent, but physically different forms:
  1. E = sigma Te^4 - sigma Ta^4, (photon particle model: difference of two-way gross flows)
  2. E = sigma (Te^4 - Ta^4) ~ 4 sigma Te^3 (Te - Ta), (wave model: net one-way flow)
where E is the intensity of the heat energy transferred from a blackbody (emitter) of temperature Te to a blackbody (absorber) of temperature Ta smaller than Te, and sigma is a constant.

Version 1 is the basis of CO2 alarmism based on "backradiation" of sigma Ta^4 from absorber to emitter, as transfer of heat energy from cold to warm.

In Slaying the Sky Dragon and Mathematical Physics of Blackbody Radiation I present a derivation of Version 2 based on a principle of finite precision computation in a wave model, without backradiation. And without backradiation CO2 alarmism crumbles.

The original version by Stefan and Boltzmann is formulated with Ta = 0 as Version 0. without backradiation (in which case 1. and 2. look identical), as an integrated version of Planck's law based on a statistical particle model.

Which is the correct formulation? Version o, 1 or 2? Particle statistics or waves? Let's list some answers from the web supposedly reflecting scientific sources:
The list can be made much longer, but we dont find any support of 1. and backradiation. And without backradiation CO2 alarmism crumbles.

The following questions present themselves:
  • Why is 1. found only in the CO2 alarmism of IPCC, and not elsewhere?
  • Is 1. a free invention which lacks original scientific source?
  • Is 1. a form of hyper-reality for which the original is missing?
  • Is 1. a form of folklore known by everybody to be true, yet without any individual scientist claiming to have demonstrated the statement.
  • Is 1. an expression of "scientific consensus" for which no original scientific source is required?
What do you think? Is CO2 alarmism based on backradiation, folklore or real science?

15 kommentarer:

  1. Is your new SB-law (2) Lorentz invariant?

    SvaraRadera
  2. It is based on Maxwell's equations which are Lorentz invariant, so yes.

    SvaraRadera
  3. The radiation transport along a ray obeys the relationship


    Assuming the local emissivity ε is isotropic, the integral vanishes and so we have simply



    In other words, excluding the losses, the local scattering and re-emission (if existent) don’t affect the radiative transfer which rest unique an directed according to Iv, that’s according to the gradient of the photonic pressure or, if you prefer, to the gradient of the temperature.
    Thus, the backradiation is pure fantasy.

    Michele

    SvaraRadera
  4. I mistook inserting images, I’m sorry. I repost.

    The radiation transport along a ray obeys the relationship

    ”(this formula)”

    Assuming the local emissivity ε is isotropic, the integral vanishes and so we have simply

    ”(this formula)”

    In other words, excluding the losses, the local scattering and re-emission (if existent) don’t affect the radiative transfer which rest unique an directed according to Iv, that’s according to the gradient of the photonic pressure or, if you prefer, to the gradient of the temperature.
    Thus, the backradiation is pure fantasy.

    Michele

    SvaraRadera
  5. I have a question about the Johnson-law of black body radiation.

    Assume that we have two bodies A and B. They are both in vacuum and at a distance of 1 light-day, ie the distance traveled by EM-radiation in 1 day.

    At time t=0 both bodies are at 0K, but they each contain a heating element which become active at that time. A is rapidly, in a few seconds, heated to 200K and B is rapidly heated to 300K.

    Does your law claim that both bodies will send out no heat for the first day and then, on day later, when they have had time to communicate their temperature to each other, B will start sending out heat to A while A never radiates anything?

    This is a particular case of the general question of how a body will "know" when to radiate if the amount of radiation sent really depends on arbitrarily distant other bodies, while still not communicating faster than the speed of light.

    The law (1) has no problem in this respect since other bodies does not influence the amount of radiation sent out by a particular body, and the net flow is just the difference between what is sent out and what is absorbed

    SvaraRadera
  6. Good question: my present model does not answer this question since it focusses on one blackbody subject to incoming forcing, for example from another blackbody, and not two blackbodies in timedependent interaction. The absorbing body in the model reads the spectrum of the incoming forcing which tells the temperature of the source if it happens to be a blackbody. The absorbing body thus does not read the temp of the source at the source at distance, but at the arrival to the absorber. All action is local and there is no action or reading at distance. OK?

    SvaraRadera
  7. I understand that the law (2) is for a time independent situation. But the principle behind (1), which is essentially linearity of superpositions, gives an answer to this question in that setting. Similarly I would expect the basic principles behind your own derivation of (2) to give an answer to the time dependent situation as well.

    We all know that in real life there are no perfectly static heat sources, so at least a qualitative explanation of a time dependent situation is required in order to attract serious interest from a physicist.

    SvaraRadera
  8. The model allows non-periodic forcing as you suggest, but the analysis presented concerns the time periodic case. The analysis can be extended of course. But turning on and off the Sun is not realistic.

    SvaraRadera
  9. That we cannot turn off the sun is something I fully agree with, but I have not asked for anything like that. I am speaking about an ideal scenario with only two bodies in vacuum in order to make it simpler to use your methods to derive the answer. Unless your derivation can be extended, and that extension is done, physicists will never begin to consider your law as a serious thing. In order to get the interest of the physics community you will need to demonstrate general principles which can be used to mathematically derive new answers. A single formula for a single situation would just be considered as curve fitting or statistics.

    One of the basic working principles in physics is that we test a principle derived in one particular by applying it to other similar situation where we do not know the answer in advance. If we get an answer which agrees with observed reality the new principle is strengthened, if it gives the wrong answer we have learned that the principle is either incorrect or less general than we thought.

    I am saying this as friendly advice since I have seen good work done by you on FEM and have respect for your mathematical knowledge. But in order attract attention in physics you need to present more than single formulas.

    SvaraRadera
  10. I am writing a book on blackbody radiation and of course hope that some physicist will be interested. So far I have seen no engagement of physicists
    in climate science in general, neither on radiation in particular. It seems to me that physicists are focussed on elementary particle physics with no interest in macroscopic physics like e.g. atmospheric radiation, so I do not expect much.
    But the world is dynamic and so things may change.

    SvaraRadera
  11. Writing a book might be a good thing in the very long run but given how the field works I do not think that you will get any attention from working physicists by doing that. Starting out with a book will probably even work against you since it is not a peer reviewed publication.

    If you want to get attention the right thing to do is to write a few papers for some of the high level physics journals, like the Physical Review Letters, where you use your own theory to derive experimentally testable conclusions. Just giving a different interpretation of the standard formulas will not work, just like "interpretations" of quantum theory is a more or less dead area, unless you count popular science journals like New Scientist which are more interested in speculation than real science.

    Your belief that most physicists are doing particle physics is not correct. Particle and high energy physics are very visible research areas so they get the attention of the press, but e.g. solid state physics is far larger in terms of people and money, thanks to large amount of corporate research.

    The main reason that most physicists are not that interested in climate modeling is that the physics involved is considered to be completely understood at the basic level and the remaining problem just one of doing large scale numerical simulation.

    SvaraRadera
  12. Physicists are fooling themselves believing that thermodynamics is understood.
    It is not clever to fool yourself, but it is a common practice in science. The idea is that if many people fool themselves at the same time, it is less foolish. My experience of peer review includes a lot of foolishness.

    SvaraRadera
  13. Well if you can provide concrete examples where your aproach gives one prediction and the alternative theory a different answer I am sure that you will find experimental physicists who are willing to test them. If you claim that physicists are fooling themselves you must give concrete examples of why you think so. Unless your theory provides experimentally testable predictions it is dead from the start.

    In physics experiment is the undisputed king, and no amount of mathematics or philosophy will stand against an experimental result. That is what separates physics from both mathematics and politics.

    SvaraRadera
  14. @ Anonym

    It seems that you assume that the power emitted by the BB is always and solely a real power putting aside that it is complex quantity.
    The power emitted that stays constant is the apparent power and the local characteristics affect the energy transfer modifying the real and reactive powers so that what doesn’t travel remains on the place and is exchanged continuously between the electric and magnetic fields.
    There isn’t any difference with the AC power excluding the fact that the power has to be computed as the Poynting vector.

    Michele

    SvaraRadera
  15. Anonym: Perhaps I can explain in simpler terms that others might also understand.

    First note that experiments show that a gas surrounding an emitter will only start to absorb radiation when the temperature of the emitter starts to exceed that of the gas. This confirms what Claes says about the cut off frequency determining whether or not radiation is converted to thermal energy.

    You must remember that coherent EM radiation is not incoherent thermal energy. There needs to be a conversion process. For example, when x-rays penetrate a material they ionize the molecules they strike, dislodging electrons. They can only do this if they have sufficient energy (as determined by frequency) above the cut-off frequency, as determined by Wien's Displacement Law - such frequency being proportional to absolute temperature.

    So, in your example, the only warming effect that will occur will start when the frequency of the emitted radiation from the hotter body (as detected by the cooler one a day after it was emitted) exceeds the frequency of the radiation that cooler body was emitting. The radiation the cooler one sends to the warmer one will never be converted to extra thermal energy in the warmer one.

    You can do experiments with identical electric bar radiators in open air facing each other (or turned away) - just compare warming and cooling times - devise your own experiments - they will all confirm what Claes says, I bet.

    SvaraRadera