söndag 20 januari 2013

Making the Fiction of DLR Real


The existence of downwelling long wave radiation DLR from the atmosphere to the Earth surface is supposedly evidenced by radiometers such as pyrgeometers and bolometers and by direct estimation of the emissivity of the atmosphere by a formula depending on temperature and vapor pressure, in both cases with coefficients determined by calibration with some chosen reference.

A pyrgeometer determines DLR as
  • DLR = U_emf/S + sigma x T^4 with T the temperature of the instrument,
where U_emf is a voltage recorded by the pyrgeometer detector and S is a sensitivity coefficient determined by calibration and sigma is the constant of Stefan-Boltmann's radiation law.

The first direct estimation by formula was given by Brunt (1932):
  • DLR = (0.605 + 0.0048 x e^0.5)  sigma  T^4, 
with T the temperature  and e the vapor pressure 2m above the Earth surface and was followed by variations by Brutsaert (1975) and many others.

We see that in both cases DLR is determined from an ad hoc Ansatz of a certain form with certain coefficients to be determined by certain calibration. We understand that the Ansatz defines DLR ad hoc by formula and calibration, but the physical nature of DLR remains hidden since the Ansatz is ad hoc without physical basis. 

Confusion arises when DLR by its suggestive name is given the physical nature of a flux of energy from atmosphere to the Earth surface. The greenhouse effect GHE is based on this confusion which magically transforms the fiction of DLR into reality.

Note that both formulas include Stefan-Boltzmann laws only valid in a surrounding of 0 K, which is not the case, and thus are ad hoc False Stefan-Boltzmann laws without physical meaning.

25 kommentarer:

  1. The LP PIRG 01 manual
    http://www.deltaohm.com/ver2012/download/LP_Pirg_01_M_uk.pdf
    as says about the device working principle:

    Radiant energy is absorbed/radiated from the surface of the blackened thermopile, creating a temperature difference between the centre of the thermopile (hot junction) and the body of pyrgeometer (cold junction). The temperature difference between hot and cold junction is converted into Potential Difference thanks to the Seebeck effect.

    Then the device works because of the ΔT produced by the heat power absorbed/radiated.

    If the pyrgeometer temperature is higher than the radiant temperature of the portion of sky framed by the pyrgeometer, the thermopile will irradiate energy and the output signal will be negative (typical situation of clear sky) vice versa if the pyrgeometer temperature is lower than that portion of sky framed, the signal will be positive (typical situation of cloudy sky).

    In other words the heat power transfer is one way. All is right, there is no fraud. The fraud is made by those who confuse the device upward radiation with the sky downward radiation.

    Michael

    SvaraRadera
  2. There is no two-way flux of heat back and forth.
    DLR is fiction.

    SvaraRadera
  3. Could you please make a precise definition of exactly what you mean by one-way heat flux and two-way heat flux.

    SvaraRadera
  4. One way heat flux is transfer of heat energy from warm to cold body, like a one-way street with all cars moving in one direction. Two-way heat flux is gross transfer of heat energy from warm to cold and from cold to warm (DLR) at a the same time, like a two-way street with cars moving in both directions. Is this clear?

    SvaraRadera
  5. No it is not clear.

    First, how do you define heat energy?

    And secondly, is there any other kinds of energy than heat involved?

    SvaraRadera
  6. Heat energy is in thermodynamics the total energy (which is conserved) minus kinetic and potential and chemical energy. It is also called internal energy, because it has a limited capability of being transformed into e.g. kinetic energy.

    SvaraRadera
  7. How does the energy of an electromagnetic wave fit into this?

    SvaraRadera
  8. I suggest a model for matter-electromagnetic wave interaction in my blackbody which carries an answer.
    Why not ask a physicist about matter-EM wave interaction?

    SvaraRadera
  9. Well, I'm a bit acquainted with EM-theory, I have read Jackson for instance. That's not important, I'm curious of your treatment.

    Does the waves leaving a body carry momentum and energy?

    It is very obscure, and you are being vague here to.

    SvaraRadera
  10. You have to look at the model I study in the blackbody book, with radiative heat transfer as a resonance phenomenon with high-frequency cut-off, and not as a particle phenomenon with gigantic infrared photons crawling in and out. Why not ask physicist, to see if that helps. They are paid to have answers.

    SvaraRadera
  11. As I said, your model is very obscure.

    How do you define a blackbody for instance?

    Further, the electromagnetic theory is not about photons, it is about waves. According to you, does these waves carry momentum and energy?

    SvaraRadera
  12. Electromagnetic waves are solutions to Maxwell's wave equation. That is all there is to say. The true nature of electromagnetism is unknown.

    SvaraRadera
  13. Does that mean that you are not sure if electromagnetic waves carry energy and momentum?

    SvaraRadera
  14. If you formulate your question as a question about solutions of Maxwell's equations it may be answered. So what is your question then?

    SvaraRadera
  15. You have an electromagnetic field F that satisfies Maxwell's equations (d*F = 4pi*J, div F = 4pi J), J is the 4-current.

    Does F carry energy and momentum?

    SvaraRadera
  16. What does "carry" mean, what is "energy" and what is "momentum" in terms of F?

    SvaraRadera
  17. Given the 4-vector p. Energy is the null component of p, momentum is the space-like three vector of p.

    dp/dt = qF,

    q is the charge.

    SvaraRadera
  18. That it transports energy and momentum from the source (that's the two Maxwell equations above) and deposit it according to the Lorentz force law, dp/dt = qF, wich gives the change in the energy and momentum of the receiving charge.

    SvaraRadera
  19. It does not make sense to me: what do mean by "transports" and "deposits" and "recieving charge". I am pretty tired of this conversation. Why don't you ask a physicist instead?

    SvaraRadera
  20. Please sir, you asked me to specify my original question.

    How does the energy of an electromagnetic wave fit into this?

    And then started this silly game of "20 questions", what do you mean with this, what do you mean with that. Why are you playing silly games?

    You seem extremely reluctant in explaining the details of your model and your train of thoughts. You say you model physics, but you do not want to discuss the physics.

    SvaraRadera
  21. I am a mathematician and I analyze mathematical models of physical phenomena and want to keep the discussion connected to the model in order to know what to say. If you want to speak about physics without precise mathematical models, I suggest you turn to a physicist.

    SvaraRadera
  22. And you know nothing about the energy and momentum in electromagnetic waves in connection to your model then?

    SvaraRadera
  23. There are many books on Maxwell's equations and electromagnetic waves with lots of information. Less is known about interaction between matter and EM-waves and blackbody radiation. In fact classical continuum physics collapsed on this problem at the end of 19th century, because of the ultraviolet catastrophe. I seek to resurrect classical continuum physics using a notion of finite precision computation, because the alternative of quantum mechanics cannot handle macroscopic pbenomena such as blackbody radiation.

    SvaraRadera
  24. Maybe you should take a closer look at solid state physics...

    SvaraRadera