onsdag 2 november 2011

Generic Model of Resonant Interaction Matter-Wave


  • $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. $- \gamma U_{ttt}$: radiation pressure from outgoing radiation = emission
  3. $- \delta^2U_{xxt}$: viscous force from internal heating = absorption
  4. $f$ is exterior forcing,
and $\gamma$ and $\delta^2$ are (small) positive constants connected to dissipative losses as outgoing radiation = emission and internal heating = absorption. Let me here clarify the physics behind the dissipative terms:
The corresponding dissipation energies are obtained by multiplication of the forces with $U_t$ followed by integration (by parts) in space-time over a period, assuming periodicity, to give
Recall that damping in the wave equation $U_{tt} - U_{xx} = f$ can take the from
$\nu U_{t}$ with $\nu$ a viscosity coefficient, or as above $-\gamma U_{ttt}$ and $- \delta^2U_{xxt}$, thus with the following derivatives:
  • $U_{t}$
  • $U_{ttt}$
  • $U_{xxt}$
where $U_t$ appears in acoustic viscous damping, and can be seen as a simplified version of both $U_{ttt}$ and $U_{xxt}$.

The above model thus can be viewed as the generic model of wave motion with damping describing of interaction matter (string) and electromagnetic waves (radiation term and forcing) with a very rich area of application including:
The mathematical analysis exhibits a fundamental phenomenon of near resonance in wave motion with small damping, as explained in How to Push a Swing: In near resonance the forcing is in phase with the displacement, while in perfect resonance (with large damping) the forcing is in phase with the velocity. The effect is that in near resonance the exterior forcing is balanced by both the string force and the Abraham-Lorentz radiation force, while in perfect resonance only by the radiation force.

Near resonance appears when the radiative damping is small ($\gamma$ is small), which is the real case of radiation with full interaction between matter (string) and radiation.

93 kommentarer:

  1. I don’t understand why you ignore that the oscillator has two simultaneous counteracting forces, i.e., a damping force μu’ due to viscosity and an emitting force γu’’’ and it simply dissipates energy if μu’ ≥ γu’’’ that’s if ω² ≤ μ/γ because the produced power is lesser than the lost one.
    The oscillator will radiate only if ω² ≥ μ/γ.
    Michele

    SvaraRadera
  2. Ok, so I have a question for you. It doesn't concern the model you discuss here but more generally the things you discuss.

    You have earlier claimed that there is a discrepancy between what you call the true Stephan-Boltzmann and the false Stephan-Boltzmann where you claim that there is a difference in interpreting it with or without using the distributive law of algebra.

    I think we all can agree that physical quantities are not dependent on what units they are expressed in.

    What is your interpretation of this if we go to a unit system in which $\sigma = 1$ (it should be around 40 in natural units).

    Sincerely,
    Dol

    SvaraRadera
  3. The model is here simplified without the switch from radiation to internal heating at a certain temperature dependent frequency. The emphasis is on the different action of the different damping terms with the damping with derivatives in space representing internal heating.

    SvaraRadera
  4. Let me clarify the argument.

    The physical laws should not depend on what unit system is applied.

    If we go to a unit system where $\sigma=1$ then we have the Stephan-Boltzmann law

    $q = T^4_1 - T^4_2$,

    and we get a frame where the so called false Boltzmann is equivalent with the so called true Boltzmann. This should then be true independently of choise of units and must then be true in all different unit systems.

    That is false Boltzmann is equivalent with true Boltzmann.

    Sincerely,
    Dol

    SvaraRadera
  5. You are missing the point, the difference between two way (false) and one way (true) transfer of heat energy.

    SvaraRadera
  6. And this couldn't imply two way transfer of radiative energy then?

    To be honest, what I have seen in the two textbooks on heat transfer I am reading at the moment there seems to be strong support for the view of two way net energy transfer.

    Sincerely,
    Dol

    SvaraRadera
  7. Start with Incropera et al, Fundamentals of Heat and Mass Transfer, ISBN 0471457280 chapter 12 Radiation: Processes and Properties and chapter 13 Radiation Exchange Between Surfaces.

    And for cementafriend, the authors are engineers (as many physicists are) ;-).

    Sincerely,
    Dol

    SvaraRadera
  8. If you can't get the book I mentioned right away and feel eager to look in to the matter of radiant heat transfer I found something interesting for you. Take a look at

    A Heat Transfer Textbook

    which can be downloaded for free. Page 31-32 (41-42 in the pdf) gives the complete coverage on how radiant heat exchange is looked upon in the physics and engineering communities. The previous pages also covers the Stefan-Boltzmann law as an empirical law that only relates to a single radiation emitter.

    Sincerely,
    Dol

    SvaraRadera
  9. Only Q net is given just as I say.

    SvaraRadera
  10. Claes wrote:
    Only Q net is given just as I say.

    On page 31 in the section Radiant heat exchange. one can read:

    Suppose that a heated object (1 fig. 1.16 a) radiates only to some other object $(2)$ and that both are thermally black. All heat leaving object $1$ arrives at objects $2$, and all heat arriving at object $1$ comes from object $2$. Thus, the net heat transferred from object $1$ to object $2$, $Q_{net}, is the difference between $Q_{1 \; to \; 2}=A_1 e_b(T_1)$ and $Q_{2\; to \; 1} = A_1 e_1(T_2)$.

    $Q_{net} = A_1 e_1(T_1) - A_1 e_b(T_2)=A_1 \sigma (T^4_1 - T^4_2)$

    Sincerely,
    Dol

    SvaraRadera
  11. Still Q net. The rest is just words with no Physics.

    SvaraRadera
  12. Ouch, miss one dollar and it doesn't look pretty :)

    Repost:

    Claes wrote:
    Only Q net is given just as I say.

    On page 31 in the section Radiant heat exchange. one can read:

    Suppose that a heated object (1 fig. 1.16 a) radiates only to some other object $(2)$ and that both are thermally black. All heat leaving object $1$ arrives at objects $2$, and all heat arriving at object $1$ comes from object $2$. Thus, the net heat transferred from object $1$ to object $2$, $Q_{net}$, is the difference between $Q_{1 \; to \; 2}=A_1 e_b(T_1)$ and $Q_{2\; to \; 1} = A_1 e_1(T_2)$.

    $Q_{net} = A_1 e_1(T_1) - A_1 e_b(T_2)=A_1 \sigma (T^4_1 - T^4_2)$

    Sincerely,
    Dol

    SvaraRadera
  13. My earlier claim was that in textbooks, in the field of heat transfer the view seems to be a two way process (net heat transfer = heat from object 1 to object 2 minus heat from object 2 to object 1).

    This supports my claim.

    I didn't express any view if it is reality or not.

    Sincerely,
    Dol

    SvaraRadera
  14. Text books are orten wrong. What is your original scientifc source?

    SvaraRadera
  15. Well, I personal view what scientific development copes well with the ideas of Popper and Kuhn.

    From this view, what is presented in textbooks and propagated from mouth to mouth among scientists in the field and whats transmitted to students are to be considered the current paradigm.

    What I present for you here seems to be the current paradigm. You are in your full right to question it and do your effort to prove it wrong, and you should try it, because it is wrong in the sence that there is no Theory Of Everything in physics (and according to my own personal belief, it never will be).

    But ask your self the question, where is the burdon of proof? It's not my task to defend the current paradigm in presenting original scientific sources, it is your task to proove that the current paradigm leads to absurd consequences. Especially since your claim is that this is the case.

    Sincerely,
    Dol

    SvaraRadera
  16. Your textbook essentially presents the view I consider correct, that is one way net flow. You make your own interpretation of two way transfer from a line in the book intended to help the innocent reader to realize that one way net flow is the bottom line. Read the book as a scientist and not as a layer seeking to twist facts to fit a certain preconceived case.

    SvaraRadera
  17. Incropera also presents one way net transfer. For an engineer this is the only thing of interest. Two way gross flow serves no purpose and lacks physical reality. Pure fiction without meaning.

    SvaraRadera
  18. When will you address the claim that your theory of one way transfer breaks down due to the bounded speed of information?

    SvaraRadera
  19. Who said it breaks down and why?

    SvaraRadera
  20. Claes wrote:
    Two way gross flow serves no purpose and lacks physical reality.

    In the case of radiation, what is your governing equation that you draw this conclusion from?

    Sincerely,
    Dol
    Dol

    SvaraRadera
  21. Those who claim two way gross flow should present their equations, not me.

    SvaraRadera
  22. Dol, I see you mentioned cementafriend in your 4th comment. I have downloaded the book and had a quick glance. Firstly, the book is written by two mechanical engineers one of whom is at MIT. The book at least mentions the work of Prof Hoyt Hottel who was the head of Chemical Engineering at MIT. I have found that many engineers including mechanicals have difficulties with chemistry and this by interference means that they are likely to have difficulties with Chemical Engineering subjects. The authors certainly appear to be confused about photons, waves and frequency. While the authors accept that increased CO2 will cause little change (from Hottels work) they appear to have swallowed the proposition that CH4 and other gases could make important contribution and provide no evidence. They should at least have looked at the section on heat and mass transfer in Perry's chemical Engineering Handbook -Prof Hoyt Hottel was one of the authors. Also have a look at my post my clicking on my name.

    SvaraRadera
  23. "Who said it breaks down and why?"

    I believe that this is the first place it as asked on your blog
    http://claesjohnson.blogspot.com/2011/08/what-judy-curry-suddenly-understands.html?showComment=1313326598418#c5970330516619924019

    You have consistently refused to provide a mathematical answer, in properly derived equations, to this question, or possibly ignored it. Instead you have given analogies, which are not proper mathematics, as you have yourself later pointed out when others have posed imprecise questions for you.

    SvaraRadera
  24. The two way flow of energy that you bizarrely believe should not exist is directly observed. To persist with your theories in the face of directly contradictory evidence is not science, it's fantasy.

    SvaraRadera
  25. @ Anonym
    The two way flow of energy that you bizarrely believe could exist isn’t directly observed. To persist with your theory in the face of directly contradictory evidence is not science, it's fantasy.
    Michele

    SvaraRadera
  26. A laymans thought...

    Imagine two poles with different levels of EMFś (electro motive forces). Both contained within a conductive surrounding. The respective EMFś are measured in relation to the common envelope of the whole setup.

    Now.. Will we really observe electrons moving in both directions between the poles? I.e two gross flows opposing each other between the poles, resulting in a correct net flow depending of the difference in their EMF levels?

    SvaraRadera
  27. Claes said:"Two way gross flow serves no purpose and lacks physical reality. Pure fiction without meaning."
    But there is a simple experiment that shows that you are wrong.
    Take a sphere with black body surfaces both outside and inside and with the same temp all over. According to your one way heat transfer theory there is zero (IR)-radiation inside because the net flow of heat is zero. 0 = 0-0 (not 0=Q-Q which is the two way heat transfer theory). If you now heat this sphere to a temp so that you can see it begin to glow, what do you expect to see inside. Total darkness, because 0 = 0-0? Drill small holes and look inside. You will see the sphere glow everywhere inside it, because 0=Q1-Q1.
    This simple experiment shows the case where the temps are equal in the SB law, with gross heat flows but net flow is zero. And one example is enough to crash your theory.
    You can extend the experiment with increasing the temp for one half of the sphere. What do you expect from that?

    SvaraRadera
  28. michele - I've directly detected radiation from the night sky, just last night. I've directly observed what you bizarrely believe should not exist. To believe your theory in preference to the evidence is bizarre indeed.

    SvaraRadera
  29. Make the experiment Lasse H and report me the result. Thought experiments are no real experiments.

    SvaraRadera
  30. Claes wrote:
    Make the experiment Lasse H and report me the result. Thought experiments are no real experiments.

    Would you acknowledge that kind of experiment as a valid one if it succeeds? For you this is really a good thing if it is, because it means that your theory is falsifiable and not subjected to be pseudoscience.

    Sincerely,
    Dol

    SvaraRadera
  31. Anonym in reply to michele, clearly you have no knowledge of instruments and errors. There is no instrument that directly measures radiation heat. The output of instruments such as thermal cameras is based on calculations & assumptions from a signal comparison such as a voltage difference. Many things can fool you such as setting of the zero, the assumption for the calculations, the equations in the calculations, the field of view, other interfering signals etc. How do you know you were not getting a signal from your body temperature? If you know about the instrument at least you should adjust the zero to zero when pointing it upwards at night in a clear sky.

    SvaraRadera
  32. The result is, your theory is wrong. Huge quantities of observational evidence show that. To persist with it is absurd. What is your motivation for making a fool of yourself?

    SvaraRadera
  33. @ Anonym

    Whatever is the device that you used, you have simply measured the heat exchanged between the device and the sky. You did the rest with your imagination and the device said what you told him to say.

    @ Lasse H

    Inside the empty sphere there isn’t electric field (that’s trivial) then no EM waves/radiation.

    Michele

    SvaraRadera
  34. Claes, do you think that it is warm inside the glowing sphere, or is it cold, or what temp is it there?
    Michele, you are wrong. The E-M-emitters are inside the sphere. You are thinking of E-M-waves coming from outside, and right, they can´t penetrate a metallic shell.
    (And, there can´t be any standing waves inside because the impedance of the E-M waves are matched to the receiver impedance, as per definition of a black body, so there is no reflection, which is a necessary condition för standing waves. You have written about this before.)

    SvaraRadera
  35. "There is no instrument that directly measures radiation heat"

    It's called a bolometer.

    "How do you know you were not getting a signal from your body temperature?"

    I wasn't standing in front of it.

    "Whatever is the device that you used, you have simply measured the heat exchanged between the device and the sky."

    No, I directly detected EM radiation emitted by the sky. I did not measure any heat exchange. The Claes Johnson theory is that the atmosphere does not radiate towards the Earth.

    SvaraRadera
  36. The statement "the atmosphere does (not) radiate towards the Earth" is ambiguous. Does "radiate towards" mean transfer of heat energy? Or does
    it mean that an electromagnetic signal can be detected by some sensitive instrument?

    SvaraRadera
  37. It is not ambiguous at all. It means both. How could a signal possibly be detected unless energy was transferred? Do you actually know any basic physics?

    SvaraRadera
  38. If it means both it is ambiguous. Which meaning are you using?

    SvaraRadera
  39. i asked 3 times the very same question, explicitly stating that i meant em-waves.
    you said it was ambigous.

    SvaraRadera
  40. I say two-way em-waves and one-way heat energy transfer. I suggest that you read and think before insisting on continued questioning.

    SvaraRadera
  41. Claes, what do you think about Lasse H's experiment? I think it sounds like a very fine experiment.

    Here are two questions for you:

    1. Do you think that the inside of the sphere will glow?

    2. If it glows, does that falsify your theory?

    I guess that your answers will be:

    1. What does "glow" mean? It is ambiguous, please provide an equation.

    2. No.

    Am I right?

    SvaraRadera
  42. Claes, have you modified your opinion? I think everyone agrees that there is two-way EM waves and one-way heat transfer. So what are we arguing about?

    SvaraRadera
  43. what happens to the em-waves emitted from the atmosphere toward the earth?
    do they bounce back?
    in that case, do they heat the atmosphere when they get back?
    how does the bouncing happen without momentum transfer?

    or do they waves go through the entire earth?
    do they heat up the atmosphere on the other side in that case?

    SvaraRadera
  44. There is a standing em-wave between atmosphere and earth which transfers energy from hot to cold by a process of absorption-emission with heating resulting from absorbed waves which cannot be re-emitted because the required coordination fails. At least according to my analysis.

    SvaraRadera
  45. Lasse H,
    The development of the Rayleigh-Jeans Law and the of the Planck law is based upon the number of standing wave modes in a cavity simulating a BB. See, e.g., http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/rayj.html#c3.

    The problem is if the EM field can exist within an empty charged sphere given that the charged spherical shield doesn’t create an electric field inside itself. See, e.g., http://www.ccmr.cornell.edu/education/ask/index.html?quid=770

    “bolometer”
    (http://en.wikipedia.org/wiki/Bolometer)
    If it has an operating temp To more lower than the lowest atmospheric temp Ta then we have Q = σ(Ta^4 – To^4) = σTa^4[1 – (To/Ta)^4] ≈ σTa^4. You are right, the effect is due only to Ta and we can assume that the device correctly measures Ta. In this case the driving physics is only the radiative heat yielded to the bolometer by the sky.
    Indeed, using a pyrgeometer (http://en.wikipedia.org/wiki/Pyrgeometer) placed at the ground (temo = Tg) and oriented upward towards the atmosphere (temp = Ta) the effect upon the sensor is due to σ(Tg^4 – Ta^4) and we simply measure the heat that it radiates towards the sky, as Tg > Ta. Then, with the software we do what we want.

    Michele

    SvaraRadera
  46. Anonym @ 8Nov8.01 Seems that you do not understand instruments note in http://en.wikipedia.org/wiki/Bolometer it clearly says that the measurement is of electrical restistance which is converted to a temperature. I repeat there is no instrument that measures radiation heat. With a cheap instrument, it does not matter where it is pointed, it will pick up some temperature reading which could be amplified. The air around a person maybe warmer than the further out surrounds. The output depends where the zero is set. The correct reading at night with an instrument in a open field at least 100m from any object (including a person)looking up at a clear sky should be zero as was found by many researchers in the past. Nobody these days bothers to read the specification and instructions on instruments but I suppose if its badly translated Chinese or Japanese there is a excuse.

    SvaraRadera
  47. that does not answer my question. i would like to know what happens to the em-waves that are emitted towards the earth when the reach the surface according to your model. they cannot be absorbed. are they reflected or are they transmitted through the earth?

    SvaraRadera
  48. They are absorbed and re-emitted without heating effect on the Earth since they originiate from a colder atmosphere.

    SvaraRadera
  49. Claes, coming back to my experiment with the glowing sphere: we both agree that there is no net energy transfer inside, but acc to your one-way heat transfer there is no (heat)radiation at all inside, which means that the temp is zero deg Kelvin (in the middle). Do you really think that? Please answer!

    SvaraRadera
  50. Ha ha, cementafriend, you really are a classic. No-one with reliable instrumentation has ever failed to detect radiation from the atmosphere when looking up at it at night. Google "infrared sky brightness".

    "They are absorbed and re-emitted without heating effect on the Earth since they originiate from a colder atmosphere."

    So, you finally admit that radiation from the atmosphere exists, and now your position is that the Earth is a perfect infrared mirror. Well, it's another absurd position that's trivially easy to prove wrong, but it's some progress anyway.

    SvaraRadera
  51. Lasse H,
    The development of the Rayleigh-Jeans Law and the of the Planck law is based upon the number of standing wave modes in a cavity simulating a BB. See, e.g., http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/rayj.html#c3.

    The problem is if the EM field can exist within an empty charged sphere given that the charged spherical shield doesn’t create an electric field inside itself. See, e.g., http://www.ccmr.cornell.edu/education/ask/index.html?quid=770

    “bolometer”
    (http://en.wikipedia.org/wiki/Bolometer)
    If it has an operating temp To more lower than the lowest atmospheric temp Ta and then we have Q = σ(Ta^4 – To^4) = σTa^4[1 – (To/Ta)^4] ≈ σTa^4. You are right, the effect is due only to Ta and we can assume that the device correctly measures Ta. In this case the driving physics is only the radiative heat yielded to the bolometer by the sky.
    Indeed, using a pyrgeometer (http://en.wikipedia.org/wiki/Pyrgeometer) placed at the ground (temo = Tg) and oriented upward towards the atmosphere (temp = Ta) the effect upon the sensor is due to σ(Tg^4 – Ta^4) and we simply measure the heat that it radiates towards the sky, as Tg > Ta. Then, with the software we do what we want.

    Michele

    SvaraRadera
  52. Michelle,every charged particle inside the shell act as a transmitter of IR-waves due to the heat (they are oscillating), so the existance of an EM-field inside the shell can´t be questioned.

    SvaraRadera
  53. Lasse,
    really,that's true if the shell isn't empty. In this case the shell should act as a burning glass and the energy density would be infinite at its centre.
    Michele

    SvaraRadera
  54. when you say that there is no heating effect since they originate from a colder source do you mean that they have the wrong frequency or that they carry some sort of label of their origin?

    SvaraRadera
  55. They have too low frequency, below the cut-off of the warmer body.

    SvaraRadera
  56. i still do not understand how it works. how can they be absorbed and emitted from the warmer body if the frequency is below the cut-off?

    SvaraRadera
  57. By near resonance. Look at the model and the analysis.

    SvaraRadera
  58. Claes, I agree with you that a colder body can`t warm a hotter one because its IR-spectrum is below the cut off freq of the hotter body. Only frequencies above the cut off freq can warm a body. Therefore only IR from earth with frequencies above cut off freq for the armosphere can warm the atmosphere. The rest is absorbed and emitted.
    But I don´t agree with you that the radiation from a black body is dependent of the background temp. That will lead to absurd (unreasonable) results. Just the net radiation is (the SB two way transfer).

    SvaraRadera
  59. Michele, thanks for the very interesting link: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/rayj.html#c3.

    SvaraRadera
  60. Thanks Lasse. I don’t agree with you about the two way transfer of the radiation. The macroscopic form of gas energy is due i) to the molecules as Pm = Nm*kT, where Pm = molecular pressure (N/m² = J/m³), Nm = molecular density (molecules/m³); ii) to photons as Pf = Nf*hv = (σ/c)T^4, where Pf = photonic pressure (N/m² = J/m³), Nf = photonic density (photons/m³).
    All the physical process occur because of a pressure gradient (P1-P2)/L. If P2 ≠ 0 then there is done a work and so an energy is transferred and this work/transfer is simply one way. If P2 = 0 the P1 is simply and totally dissipated without any useful effect.

    Michele

    SvaraRadera
  61. Michele, take a case with two blackbody surfaces, both with temp T, exchanging energy (suppose empty space between them). I say that both emit energy acc to R=sigmaxTexp4 and both absorb the same energy. i.e.the net energy exchanged is zero. If I understand you right you say that the surfaces emit zero and absorb zero, i.e there is no radiation between the surfaces. Do I understand you right?

    SvaraRadera
  62. Lasse, you have understood me very well. The constructive and destructive interference within the space between the two BB surfaces allows that only the standing waves survive whereas all the others vanish. The energy doesn’t travel and it is simply stored and maintained on the place. Then no energy travel no energy transfer.
    In the simple case of unidirectional counter traveling waves we have
    Asin(kx+ωt) + Asin(kx-ωt+φ) = 2Asin(kx+φ/2)cos(ωt+φ/2)
    (prosthaphaeresis formulas at http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Product-to-sum_and_sum-to-product_identities)
    That’s, we have always a standing wave whatever is the phase between the two counter traveling waves.

    Michele

    SvaraRadera
  63. P.S.
    Then, no traveling wave no energy transfer no emission no absorption.
    Michele

    SvaraRadera
  64. So, acc to you (and Claes), there is zero deg K between the BB surfaces because of no energy transport. This is an absurd consequence of your theory. Suppose the temp T of the surfaces is 1000 deg K, then I don´t think you would like to be there in between. I think you will be heated up to that temp. And suppose you can stand it (you are still alive), acc to you, you will not see the glowing surfaces. Absurd, isn`t it?
    With the two way energy transfer, there will be no such absurd results.

    SvaraRadera
  65. Lasse, if the inner space is empty then saying of temp is a non sense as the temp is the macroscopic effect of the molecular kinetic energy or in general of the material energy carriers. Assuming the space is filled with a gas it would be isothermal and then isoenergetic because of the equipartition of kinetic energy. The gas must have the same temp of the surfaces. If the space is empty it continues to be isoenergetic as the EM energy too is equipartitioned. But, in both the cases there isn’t any energy transfer.
    Of course, if we introduce anything in that space the new system will reach a new equilibrium, if needed. Once the transient is finished any energy transfer will stop again.

    Michele

    SvaraRadera
  66. Michele, to be able to talk about temperature, of course there must be some kind of material in the space, e.g a thermometer. Introducing gas will not give a pure radiation case, because then you will have conduction and convection phenomena. I still claim acc to your one way heat transport theory: you can´t see the glowing surfaces if you are looking inside there, because acc to you, the surfaces are emitting nothing (and absorbing nothing) . If you measure the temp there, in a space without radiation there will be zero deg K.
    I say, they are emitting acc to the SB law R=sigmaxTexp4 and absobing the same. We probably won´t come any further?

    SvaraRadera
  67. Lasse, the inner space of the cavity isn’t empty but it is filled with standing EM energy. A thermometer introduced in the middle of that space will measure the surfaces temperature even if it is not placed in contact with them.

    Michele

    SvaraRadera
  68. a standing wave needs (at least) two travelling waves and i still do not understand how the warmer body can emit and absorb at frequency that are forbidden to it by the sharp cut-off that clas's model requires.
    i do not understand the analogy with the swing since the frequencies from the colder body can be very far from the cut off, they are not (at least not all of them) near resonance.

    SvaraRadera
  69. lorenzo, as I see it: the frequencies of the IR-spectrum from a colder body lies below the cut-off freq of a warmer body and can thus be absorbed (and reemitted) by it, thus not warming it. Only frequencies above the cut off freq can warm a body, e.g. the visible radiation from sun, warming the earth. In the same way, only IR-frequencies from earth above the cut-off of the atmosphere can warm the atmosphere, if they can be absorbed, otherwise they are just passing through.

    SvaraRadera
  70. ok i see the point now, thanks.

    SvaraRadera
  71. Michele, there can´t be any standing waves unless the surfaces are emitting radiation. Otherwise the standing waves will disappear.

    SvaraRadera
  72. Lasse, the inner space between the two opposite surfaces is filled with two counter traveling waves for any frequency having respectively the amplitude A and B, both function of temp. If A>B (i.e., Ta>Tb) then the global effect is due to the standing wave of amplitude 2B plus the traveling wave (A-B) directed from A to B.
    The standing wave doesn’t exist only if there is missing the surface A or B. If the two surfaces exist simultaneously then the standing wave exists too. And if A=B then the traveling wave vanishes and there exist only the standing wave, that’s there is no heat transfer, then no emission nor absorption.

    Michele

    SvaraRadera
  73. Michele,we are talking about the case Ta=Tb. A standing wave can´t exist just by itself, without emission from any of the surfaces A or B. Maybe in an ideal case with total reflection of the waves at both A and B, but a BB does not reflect the waves, thats why it is called a BB: it absorbs (and emits)radiation with proper frequencies. So,sorry, there can´t be any standing waves in the case Ta=Tb, unless the surfaces are emitting.
    And the surfaces emit (and) absorb acc to R=sigmaxTexp4.

    SvaraRadera
  74. Lasse, the standing wave due to A exists because i) it has been firstly emitted by A ii) it has reached B iii) it cannot continue its travel as A has blocked it iiii) the space between A and B remains filled with the energy emitted by A. That is also valid for the standing wave due to B. Thus, the standing wave exists because two waves have been respectively emitted by A and B and they remain blocked in the space.
    The EM wave acts as a pipeline (I think it is an EM pipeline) which has to be firstly filled, e.g. with water, and then it is able to transfer water from the reservoir A to the reservoir B and, if the difference of the total pressures at its ends is zero, it simply remains filled with standing water and there doesn’t exist any water flow.
    The physics is the same if the energy carriers are molecules or photons.
    Michele

    SvaraRadera
  75. Michele, the standing waves can continue to exist only if they are reflected at both surfaces, as I mentioned before, but in this case with BB surfaces, they are not reflected because a BB surface absorbs incoming waves and reemits it, e.g. you will have emission at both surfaces even if they have the same T. But of course these waves can interfere with each other as you have described before. But still, the surfaces are emitting acc to the SB law!

    SvaraRadera
  76. Lasse, the standing waves between A and B have a node both on A and on B. So δE/δt = δH/δt = 0 and there isn’t any interaction wave/matter, then no emission no absorption.
    Michele

    SvaraRadera
  77. ...And still the surfaces are glowing, when you look at them through peeping holes. What kind of emission is that? Or do you think you will see darkness? What you describe is the phenomenon "reflexion", the action of a mirror, and a BB is not a mirror, because absrption and emission factors both =1.

    SvaraRadera
  78. Lasse, You look the exiting radiation not the internals walls of the cavity. The radiation exit because of the peeping hole which breaks the equilibrium of EM field obtained with the standing waves. Closing the hole the equilibrium is restored. Within the cavity in equilibrium ( not communicating with the outside) there aren’t reflection as the EM fiend is lacking at the walls.

    Michele

    SvaraRadera
  79. michele what's the origin of the nodes in A and B?

    SvaraRadera
  80. Lorenzo, the waves having the node on A and B are the sole which allow the equipartition of energy within the empty space amid that surfaces without energy transfer, that’s, the equilibrium of the energy density.

    Michele

    SvaraRadera
  81. Michele, a peeping hole with micro dimensions compared to the infinite surfaces can´t affect any equilibrium.
    1) so you claim that the surfaces inside, with a temp T about 1000 deg C are dark, if you could see them (= they are not emitting)?
    2) If the temp of one of the surfaces increases with 100, 200, 300,..., 1000 deg, what will happen? Will you eventually see the hotter one start to glow? And the colder one, at 1000 deg, is still black?
    3) what do you mean by "fiend"?

    SvaraRadera
  82. Lasse, a peeping hole with micro dimensions compared to wavelength of radiation a duct with an infinite diameter and it affect the internal equilibrium: the radiative pressure decreases and the surfaces emit what is enough to restoring it.
    The temperature sensor sees the surfaces if they are hotter as it always measures/feels/sees a difference of temp..
    Sorry, “fiend” is a typo. I intended “field”.
    Michele

    SvaraRadera
  83. Lorenzo, allow me to integrate my previous post. I suggest a glance to Electromagnetic Waves in a Cubical Cavity at http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/rayj.html#c2" where we read “The solution to the wave equation must give zero amplitude at the walls, since a non-zero value would dissipate energy and violate our supposition of equilibrium. To form a standing wave, the reflection path around the cavity must produce a closed path.”
    Michele

    SvaraRadera
  84. Michele said: "the standing waves between A and B have a node both on A and on B. So δE/δt = δH/δt = 0 and there isn’t any interaction wave/matter, then no emission no absorption"

    The statement, δE/δt = δH/δt = 0, is not true because that is valid at Emax/min and Hmax/min.
    Instead E=H=0 at nodes. But this perhaps doesn´t matter?
    Try to answer question 2) in my earlier enter.

    SvaraRadera
  85. Lasse, you are right I mistook; at the wall we have E=H=0 as there are E and H to doing work on the electric charges of the wall. Thanks.
    About your question 2).
    a) If Ta=Tb, A and B are both black.
    b) If Ta>Tb, A becomes glowing and B is still black.
    All that is implicit in my claim “The temperature sensor sees the surfaces if they are hotter as it always measures/feels/sees a difference of temp.

    Michele

    SvaraRadera
  86. Michele,to me it seems a little strange that a BB surface at 1000 deg C is black and that a BB surface at 2000 deg C is just glowing.
    I look at the BB radiation like this: at the surface there are (uncountable) EM transmitters/receivers (in the same body) in form of molecules, charged in some way, e.g. dipoles or ions. These are vibrating at certain frequencies depending on temp T of the sutface. The frequencies show up in the IR-spectrum from the BB. My question is: how could IR-radiation from another BB get the molecules to stop the radiation, in spite they still are vibrating as before (T is constant). This is a riddle to me.

    SvaraRadera
  87. Michele, said: "All that is implicit in my claim “The temperature sensor sees the surfaces if they are hotter as it always measures/feels/sees a difference of temp."
    But a frequency sensor/counter is not temperature dependent,e.g. your eye(s), it doesn´t know what temp the source has. A light source can be cold or warm (LED or hot BB), but the frequency can be the same.

    SvaraRadera
  88. Lasse, you are right about the frequency sensor (I was think about a thermal sensor) but this confirms what I was claiming: a body can/cannot glow regardless of its temp. Then, why the glowing?
    I think to read well the contribute of the electrons to radiating energy exchange at the external wall of the BB.
    1) The BB wall emits/absorbs the energy adducted/abducted by the free electrons which play the part of heat carriers by means of their kinetic energy.
    2) Close to the wall the electron needs to reverse the direction of its motion because it has to return inside the BB and doing that it accelerates/decelerates if it is absorbing/emitting energy, that’s, if it is warming/cooling.
    3) The electron energy increases/decreases if the electric field of the EM wave closest to the wall is working on the electron or vice versa.
    4) The decelerating electron radiates both by bremsstrahlung (tangent deceleration) and by synchrotron radiation (centripetal acceleration).
    5) Within the BB in thermal equilibrium no heat carriers go back and forth, then, no exchange of energy occurs at the wall (no emission and no absorption)
    6) The electron cannot accelerate and decelerate simultaneously and it changes its energy only one way and then the radiation on the wall is only one way.
    I think my reasoning is very coherent and rational. I don’t read any enigma in that.

    Michele

    SvaraRadera
  89. Sorry Michele, your arguments don´t convince me that your theories are right. If you can show that they are, you will get the Nobel Prize.
    I don´t understand what you mean by:"a body can/cannot glow regardless of its temp. Then, why the glowing?"
    I suppose you don´t agree with the following, copied from Wikipedia (link at the end):
    "the formulas below shows that a human, having roughly 2 square meter in surface area, and a temperature of about 307 K, continuously radiates approximately 1000 watts. However, if people are indoors, surrounded by surfaces at 296 K, they receive back about 900 watts from the wall, ceiling, and other surroundings, so the net loss is only about 100 watts."
    http://en.wikipedia.org/wiki/Thermal_radiation

    SvaraRadera
  90. Lasse, I see you want to joke but thank you for your wishes, all should occur (ha! haaa!).
    The mine is only a point of view which takes into account also the SW radiation of BB that your point of view don’t assess because the lattice bounds can be at least the cause of the IR or better of the FIR. But the BB spectrum has both the very short wavelengths and the very long ones.
    After all, one don’t spend anything to thinking.
    "a body can/cannot glow regardless of its temp. Then, why the glowing?"
    The glowing isn’t caused by the temp.
    Wiki link
    Wikipedia isn’t the Bible! The two way radiation is a pure imagination. I must confirm what I said before: the electric charge emits EM decreasing its own kinetic energy (decelerating), or it absorbs EM increasing its own kinetic energy (accelerating). It cannot do that simultaneously. Thinking that is out of any logic.

    Michele

    SvaraRadera
  91. Michele said:"The glowing isn’t caused by the temp." No, but there is a clear connection between them, the frequency spectrum of a BB as a function of T.
    M said:"Thinking that is out of any logic."
    But with your (and Claes´) one way radiation theory, you get black BB surfaces at 1000 deg C, which should be a mystery to you.

    SvaraRadera