A non-relativistic derivation of Eo=mc^2 and the inertial mass of a particle

Discussion in 'Alternative Theories' started by Richard Gauthier, Dec 27, 2016.

  1. Richard Gauthier Registered Member

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    Dear rpenner,
    Academia.edu, researchgate.net and other online sites publish non-peer-reviewed articles. That's great! It's very difficult to get an "alternative physics" article peer reviewed and published by a standard physics journal or even on arXiv. But why would I waste my time self-publishing hypotheses that are "fraudulent, knowingly false, misleading or deceptive". Please cite any evidence that I claim that any of my hypotheses are on "firm theoretical ground". I'm proposing here a new concept, a spin 1/2 charged photon, that may compose an electron and other particles. Can you prove that such a variety of photon (as I choose to call it) does not physically exist? I didn't think so. And by the way, my article "Electrons are spin 1/2 charged photons generating the de Broglie wavelength" is published in the online proceedings of the San Diego SPIE photonics and optics conference of August 2015. This gives a date marker for this hypothesis as can be seen on the article at https://www.academia.edu/15686831/E..._photons_generating_the_de_Broglie_wavelength . What you call "aping the form of physics papers" is actually formatting articles in an accepted scientific format, which is the standard way of formatting original physics articles and is a prerequisite for having an article published in a peer-reviewed journal, as you know. You could be a little more courteous with your language. I do very much appreciate the attention and critical thought and thoughtful comments that you have given to these ideas, however.
     
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  3. Richard Gauthier Registered Member

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    Hello rpenner,
    This is a fairly comprehensive summary of my work. Thank you. I didn't know about the Dissident Scientists citation. I've focused on the electron because I early on found (early '90s) that a simple electron model could be composed of a curled up double-looping photon model. My idea was first published in 1992 in a short magazine article that you may have missed, "The new science of microvita" at https://www.academia.edu/20064515/The_New_Science_of_Microvita . It was written when I was a yogic monk living in Europe. Other researchers have independently come up with this idea, such as Williamson and van der Mark in their 1997 article "Is the electron a photon with toroidal topology?" at http://home.claranet.nl/users/benschop/electron.pdf . So in this view an electron may be less "fundamental" than a photon. I recently discovered that the inertial mass m=Eo/c^2 of a fundamental particle like an electron can be derived non-relativistically using F=dp/dt =ma from the rotating momentum vector in a model a resting electron composed of a circling photon-like object. Isn't that interesting? This unexpected result was an outcome of my previous electron-modeling work. My finding of a new interpretation of the relativistic-energy momentum equation in terms of the possible momentum structure of an electron composed of a helically-moving charged photon-like object was another interesting and unexpected outcome of my earlier work. There are also some preliminary indications that my electron-modeling approach may give some insight into understanding why the Schroedinger equation works, at https://www.academia.edu/10235164/T...of_the_Electron_Fits_the_Schrödinger_Equation . So I'm optimistic that this work may lead to some modest advances in physical understanding.
     
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  5. James R Just this guy, you know? Staff Member

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    Richard Gauthier:

    You posit that the momentum of your photon-like circling object has momentum related to its kinetic energy by \(p=E/c\).

    I am wondering: is this an assumption, or a result that you can derive in some other way (other than by using relativity, of course)?
     
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  7. Richard Gauthier Registered Member

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    Hello James R,
    Thanks for your question.
    It's an experimental fact that the momentum p of a photon is equal to its energy E=hf divided by c: p=E/c = hf/c . No need to derive it. My photon-like object has this property of a photon. And the energy E of a photon is not called its kinetic energy. That term is reserved for particles with mass. KE = (gamma-1)mc^2 for relativistic particles with mass, or approximately KE = 1/2 mv^2 for low velocities.
    Richard
     
  8. origin Heading towards oblivion Valued Senior Member

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    11,888
    You have been persuing this for like 25 years? Geeze what a waste of time.
    Well, I guess it is no worse than any other hobby and probably costs less.

    Please Register or Log in to view the hidden image!

    So enjoy your mental gymnastics, I just hope you aren't fooling yoursef into thinking that you have a viable theory that will lead anywhere.
     
  9. Richard Gauthier Registered Member

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    Hi Origin,
    Thanks for your encouragement. Happy new year!
     
  10. Richard Gauthier Registered Member

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  11. origin Heading towards oblivion Valued Senior Member

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    11,888
    You too! I hope we all survive the first year of a Trump administration....
     
  12. origin Heading towards oblivion Valued Senior Member

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    11,888
    That was not very kind of you to lump exchemist and rpenner in with farsight!
     
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  13. Richard Gauthier Registered Member

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    Origin,
    I didn't lump them together. They are all part of this forum. Have a Trump year.
    Richard
     
  14. rpenner Fully Wired Valued Senior Member

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    It's experimental fact that the relation between energy and momentum of a photon is consistent with prediction of electromagnetism E = c | p | which in turn is consistent with the theory of massless particles in the intersection of quantum mechanics and special relativity. But in all those examples, momentum is strictly conserved and the photons move in straight lines. Therefore you cannot derive p = E/c from these theories or a strong analogy with a photon.

    But even the best experiment can't prove a relation like the above. The currently accepted experimental upper bound on a photon's mass is \(10^{-18} \, \textrm{eV}\) which would say a 550 nm photon might travel slower than c by as much as 931 nm / billion years, with corresponding deviation from your summary of SR by about 1 part in \(10^{37}\). Since none of those experiments show momentum traveling in circles instead of straight lines, they do not approximately support your assumptions.
    http://pdg.lbl.gov/2016/listings/rpp2016-list-photon.pdf

    Back before we had confirmed atomic theory, in Einstein's 1905 paper Concerning an Heuristic Point of View Toward the Emission and Transformation of Light, he basically invents the photon to explain the thermodynamics of black body radiation with confirmation in the photoelectric effect, thus unifying two distinct phenomena under a single theoretical framework.
    He took Planck's experimentally successful 1901 formula for energy density at a given (cyclical) frequency:
    \(\rho_{\nu} \, d \nu \; = \; \frac{\color{red}{\alpha} \nu^3}{e^{\color{red}{\beta} \nu / T} - 1} \, d\nu\)
    And it's asymptotic behavior at low values of \(\nu /T \) is:
    \( \rho_{\nu} = \frac{\color{red}{\alpha}}{\color{red}{\beta}} \nu^2 T - \frac{1}{2} \alpha \nu^3 + \frac{1}{12} \alpha \beta \nu^4 / T - \dots \)
    While the a thermodynamic argument suggests that
    \( \rho_{\nu} \approx \frac{R}{N_A} \frac{8 \pi}{c^3} \nu^2 T \)
    and then hypothesizes that \( \frac{R}{N_A} \frac{8 \pi}{c^3} = \frac{\color{red}{\alpha}}{\color{red}{\beta}}\) in the low-frequency, high-temperature limit.

    Nowadays, the thermodynamic gas constant is written in terms of Boltzmann's constant and Avogadro's number : \(R= k_B N_A\).
    So \(\frac{\color{red}{\alpha}}{\color{red}{\beta}} = \frac{8 \pi k_B}{c^3}\) (Section 2)
    In sections 3, 4 and 5, he argues that the thermodynamic entropy of blackbody radiation resembles that of an ideal gas.
    Finally in section 6 he shows that the energy of an "atom" of that blackbody radiation needs to have energy \(E = \frac{R \color{red}{\beta}}{N_A} \nu = k_B \color{red}{\beta} \nu\). Inventing the symbol, h, to stand for that constant of proportionality, we have:
    \( E = h \nu \\ \color{red}{\beta} = h/ k_B \approx 4.79924466 \times 10^{-11} \, \textrm{Kelvin per Hertz} \\ \color{red}{\alpha} = \frac{8 \pi h}{c^3} \approx 6.18064462 \times 10^{-58} \, \textrm{Joules per Hertz}^4 \, \textrm{per meter}^3\\ \rho_{\nu} \, d \nu \; = \; \frac{\frac{8 \pi h }{c^3} \nu^3}{e^{\frac{h \nu }{ k_B T}} - 1} \, d\nu\)
    Which is the modern form of Planck's 1901 law.
    https://en.wikipedia.org/wiki/Planck's_law#Spectral_energy_density_form
    http://pdg.lbl.gov/2016/reviews/rpp2016-rev-phys-constants.pdf

    A neat trick, but then Einstein tries to push his analogy with particles of an ideal gas until it breaks. Because the author of an idea is the easiest one to be fooled into thinking the idea is a good one. In science, we need something more than soi disant experts.
    In section 7, he explains that Stokes's rule of photoluminescence seems naturally consistent with his model.
    https://en.wikipedia.org/wiki/Stokes_shift
    In section 8, he explains that weak source photoelectric effect seems naturally consistent with his model.
    In section 9, he characterizes the ionization of dilute gasses as something with a fixed energy scale, which corresponds to the modern view of the ionization of gases at the scale of ~ 10 eV, and makes a positive prediction.

    Here we see the difference between theory and experiment. Experiment can't say the energy of a photon is \(h \nu\), but only that it is close to that value for certain light, within experimental uncertainty. Likewise, it should be clear that no series of experiments can show that \(E = p c\) for photons, but only that the photons tested have that momentum (if tested singlely) or average momentum (if tested en masse). To say that \(E = p c\) one has to go beyond summarizing experiment and make the generalization common to all physical theories.

    A scientific theory is a communicable framework for describing precisely the observable behavior of a large class of related phenomena. The equals sign connecting physical quantities is a precise description, more precise that experiment can prove. Experiment can only show if a given test of a theory is consistent or not consistent with that theory.

    Here, we are told that (given the electron is at rest) \(E = m_e c^2\) is assumed, \(p = m_e c\) is assumed, but momentum is not conserved, resulting in a helical path through space-time. This does nothing to explain the rest mass of the electron and discards the rigid framework of SR without replacement to explain why momentum is conserved in experiments.
    You have failed to elucidate any principle.

    Nothing is wrong with defining (in SR) Kinetic energy as the difference between the energy of a body and c² times its invariant rest mass. It's consistent with your SR-derived (γ−1)mc² rule (which only applies to massive particles) as well as the equally effective \(E - \sqrt{E^2 - c^2 p^2}\) which works even for massless particles, and the E−E₀ conceptual model and naturally extends to massless bodies which carry energy and momentum.

    For a free particle, SR dictates: \(E^2 = (mc^2)^2 + (c p)^2 \) and \(E \vec{v} = c^2 \vec{p}\).
    For a free particle, QM dictates: \(E = h \nu = \hbar \omega \\ p = h / \lambda = \hbar k \)
    For a wave, we have \(v_{\textrm{phase}} = \lambda \nu = \omega / k \) while \(v_{\textrm{group}} = \frac{\partial \omega}{\partial k}\)

    Putting those together, we have \(v_{\textrm{group}} = \frac{\partial \omega}{\partial k} = \frac{\partial h \omega}{\partial h k} = \frac{\partial E}{\partial p} = \frac{\partial mc^2 \sqrt{1 + \frac{p^2}{m^2c^2}} }{\partial p} = \frac{c^2 p}{\sqrt{(mc^2)^2 + (c p)^2}} = \frac{c^2 p}{E} = v\), \(v_{\textrm{phase}} = \frac{\omega}{k} = \frac{E}{p} = \frac{c^2}{v}\).

    Plus, as momentum and energy are properties of the excitation of the wave function, there is no conceptual stumbling block to working with massless particles that convey energy and momentum across finite distances.

    That's the power of theory.
     
  15. James R Just this guy, you know? Staff Member

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    Richard Gauthier:

    The point is, it is a derivable result from special relativity, whereas apparently you need to assume it as an axiom of your theory.

    Doesn't this mean you're sneaking in some special relativity by the side door, in effect? You're claiming to have a non-relativistic derivation of \(E_0=mc^2\), but if you can only get to that by assuming a relativistic formula (\(p=E/c\)) to start with, then your derivation is not a complete non-relativistic derivation, as far as I can tell.
     
  16. Richard Gauthier Registered Member

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    Hi James R,
    I'm not saying that relativity is incorrect. But if special and general relativity had never been developed, it is still the case that experimentally, p=E/c for a photon. This experimental relationship for a photon could have been (and perhaps was?) established in a physics laboratory without using relativity (but still using the relation E=hf which is not a relationship from relativity). And this experimental fact, combined with my hypothesis of a spin-1/2 charged photon with energy Eo moving with momentum p=Eo/c in a circle to form an electron model, is sufficient to derive the inertial mass m=Eo/c^2 of the electron model from F=dp/dt = ma , where m is the electron model's inertial mass and "a" is the centripetal acceleration of the photon moving in the circle.
    Richard
     
  17. rpenner Fully Wired Valued Senior Member

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    Citation required. Physics is ruled by experiment, but theory is the basis for interpreting experiments. Without theory, all you have is a bunch of data.

    Historically, relativity informed the theoretical and experimental exploration of the momentum of the photon.
    In 1905, Einstein reexplained Planck's 1904 formula in terms of the energy of individual photons. In a separate pair of papers he developed Special Relativity.
    In that same decade, Minkowski and Poincaré gave us a geometrical formulation of SR and a theory of invariants. From this, p = E/c = hf/c = h/λ was the model for the photon.
    In 1923, Compton explained the results of shining X-rays on carbon in terms of the momentum \(h \nu/c\) of individual photons. http://journals.aps.org/pr/abstract/10.1103/PhysRev.21.483 But right there in figure 1B, relativity is used to describe the momentum of the electron. Likewise, relativity is used to describe the geometry of the scattering angle.
    This was a reversal of his published opinion from scant years before.
    In 1924, DeBroglie wrote his thesis where he associates momentum with h/λ as a general hypothesis.
    In 1925, Compton and Simon confirmed Compton's theory of individually scattered photons by measuring the recoiled electron in relation to the scattered photon.
    https://www.nobelprize.org/nobel_prizes/physics/laureates/1927/compton-lecture.pdf

    Key to all of this was a comprehensive theory of energy and momentum conservation, therefore a unified framework for associating energy and momentum with particles and totaling them up between collections of particles. That's the work of Special Relativity.

    1) Not a photon if it has electric charge.
    2) Not a photon if it has spin-1/2
    3) Not momentum if it moves in a circle.
    4) Not a derivation of inertial mass when E₀ is a free parameter which you set to the trivial value \(m_e c^2\)
    5) You neglected your second free parameter R, from which comes the purely geometric relations: ω = c/R and a = c²/R = R ω², so dp/dt = ω p = ω E₀ / c = E₀ / R.
    6) You neglected your problem with relativity in that if observer A sees an electron at rest, observers B, C, and D may each faithfully describe it as moving with speeds c/2, c/3, and c/4 in directions which are all at right angles to each other. What then is the axis of rotation of the electron? In which frames is the internal motion circular? To think that you can sweep this under the rug means that, contrary to your assertion that you're “not saying that relativity is incorrect”, you are working in a conceptual universe uninformed of special relativity.
    7) Your model of centripetal acceleration violates conservation of momentum, because ∑F≠0, which requires you to a) establish what new laws of physics are obeyed by your ur-electron and b) establish how currently accepted momentum conservation arises at the level of experimental confirmation.
    8) Unexplained is the relation between the muon and electron. If the muon is of the same type of the electron, why don't positrons and muons annihilate?
    9) You also seem to be working in a conceptual universe uninformed of quantum mechanics in that you ascribe a definite position and momentum to the ur-electron instead of treating it with wave mechanics thus claiming various things about orbital momentum.
    10) You also seem to be working in a conceptual universe uninformed of conservation of angular momentum, for if the ur-electron has intrinsic angular momentum and it has orbital angular momentum, what then is the experimentally observed angular momentum?

    You have a real problem here because physical theories (like geometry) don't just switch off when it's inconvenient to the author. While E₀ is a trivial rewriting of \(m_e c^2\), especially in systems of units where \(\hbar = c = 1\), your free parameter R is a huge, unevidenced assumption in a realm where one might expect to find some evidence.

    In 1905, atomic theory was just a hypothesis with indirect experimental support. Today we can image atoms with TEMs and grope them with STMs. Your hypothesis of 0.4 pm diameter electrons is hard to swallow in a world where the shape of 0.004 pm nuclei are studied.

    https://en.wikipedia.org/wiki/High-resolution_transmission_electron_microscopy
    https://en.wikipedia.org/wiki/Scanning_tunneling_microscope
     
    Last edited: Jan 6, 2017
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  18. Richard Gauthier Registered Member

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    Wikipedia: Photon

    In 1909[42] and 1916,[44] Einstein showed that, if Planck's law of black-body radiation is accepted, the energy quanta must also carry momentum p = h/λ, making them full-fledged particles. This photon momentum was observed experimentally[45] by Arthur Compton, for which he received the Nobel Prize in 1927.

    Compton, A. (1923). "A Quantum Theory of the Scattering of X-rays by Light Elements". Physical Review. 21 (5): 483–502. Bibcode:1923PhRv...21..483C. doi:10.1103/PhysRev.21.483.
     
  19. Richard Gauthier Registered Member

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    rpenner,
    Thanks for the further explanations. Yes, theory is very powerful.
     
  20. paddoboy Valued Senior Member

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    Let me express my respect for you Richard and congrats.....
    No, I don't agree with your hypothetical alternative, but at least you have come here, with some humility and posted in the correct section.
    That is actually like a breath of fresh air! For that reason alone, you have my respect.
    We have many agenda laden cranks, quacks, god botherers etc, that come here under different guises, by immediately trying to deride and dismiss SR/GR and many other areas of cosmology, in the sciences....
     
  21. Richard Gauthier Registered Member

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    Hello paddoboy,
    Thanks. This seems like a good place to get new ideas seriously critiqued, unfortunately by anonymous critics whose credentials are also unknown. I wonder about their own track records in producing constructive original ideas and hypotheses that could advance physics significantly if correct.
     
  22. Richard Gauthier Registered Member

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    Hello rpenner,
    You're asking a lot of excellent questions.

    Consider the case of electron-positron pair production from a sufficiently energetic photon in the presence of a nearby atomic nucleus that can absorb the excess momentum in the pair production process. QED considers this a "black box" process where a photon goes in and an electron and positron come out. But what is going on physically when a photon converts to an electron positron pair? I'm proposing that the spin 1 photon divides into two spin-1/2 charged photons (conserving angular momentum), one positive and one negative (conserving electric charge) . The positive and negatively charged spin-1/2 photons curl up to become the positron and the electron. Perhaps if electron-positron pair production is researched more closely with this hypothesis in mind, evidence of two spin-1/2 charged photons forming from a spin 1 photon and curling up to form a positron and an electron will be found.

    The "free parameter" R in the circling charged photon model of the electron is Ro= hbar/2mc which comes up in the Dirac equation solution's amplitude for a free electron. But in any case the R cancels out in the derivation when the force F= dp/dt = wp = Eo/R is divided by the centripetal acceleration a = c^2/R , giving inertial mass m = (dp/dt)/a = Eo/c^2 for a charged photon moving in a circle with any radius R. By the way, Eo = 0.511 MeV for an electron is not an arbitrary parameter. It's the resting energy of an electron. We all know that Eo=mc^2 for a resting electron. What is unknown is the ORIGIN of this inertial mass m from the resting energy Eo of an electron . Using units like c=1 does not explain anything.

    In the circling charged photon model of the electron there is a central force F=dp/dt = wp on the circling photon which is calculated to be 0.424 Newtons . So momentum is not conserved in the model because this force is changing the momentum of the circling charged photon. The cause of this force is not known. But it doesn't change the energy of the photon because it acts perpendicularly to the direction of motion of the photon. The average momentum of the circling photon in a resting electron is zero. If the relativistic energy-momentum equation is considered in terms of average quantities, the average energy of the resting electron model is Eo, the average momentum of the circling charged photon is 0 and the average mass is found from E^2=p^2 c^2 + m^2 c^4 to give Eo^2 = m^2 c^4 or m = Eo/c^2 for the average invariant mass of the circling charged photon . A circling photon has an invariant mass E/c^2 while a normal photon has 0 invariant mass.

    I showed that the relativistic energy-momentum equation for a relativistic electron model is equivalent to P^2 = p^2 +(mc)^2 where P=E/c is the total momentum of the helically moving charged photon along its helical trajectory, p=gamma mv is its longitudinal momentum component of the charged photon's total momentum (and equal to the momentum p=gamma mv of the modeled moving electron) and mc is the internally circling transverse momentum component of the helically moving charged photon's total momentum. Since the transverse momentum vector mc is always perpendicular the the longitudinal momentum vector p=gamma mv of the modeled electron (following the Pythagorean theorem evident in the momentum equation above ), it means that the longitudinal spin component of the moving electron (which is calculated partly from mc orbital momentum and partly from the helically moving spin 1/2 photon) is always directed parallel or antiparallel to the the direction of the momentum vector (and velocity vector) of the moving electron as observed by an observer in ANY moving inertial frame relative to the resting electron. The electron is observed always to have spin + 1/2 or -1/2 i.e. spin up or spin down for any observer.

    Clearly a spin 1/2 charged photon for an electron would be different from the spin 1/2 charged photon for a muon or tau. We haven't discussed this, but each would have its own rest mass equal to the mass of the electron, muon or tau.

    You are welcome to call the spin 1/2 charged photon a UR particle if you like. I prefer to call it a variety (or species if you like) of photon.

    Remember that the charged-photon electron model is not a solid mass with a particular radius. It's composed of a circling point-like charged particle moving with an extremely high orbital angular frequency given by Wzitt= 2mc^2 /hbar = 1.55 x 10^21 rad/sec due to its double-looping motion, with a circle radius of Ro=hbar/2mc = 1.93 x 10^-13 m. Its centripetal acceleration is Acent = Wzitt^2 Ro = 4.66 x 10^29 m/s^2 .
     
  23. rpenner Fully Wired Valued Senior Member

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    Einstein's 1909 far-ranging paper touches on, but does not clearly prescribe the momentum of a photon. He spends the bulk of his time killing off the ether and wave theory of radiation.

    His 1916 paper assumes it outright in the last two paragraphs of the introduction.

    If the assumed hypotheses about the interaction of matter and radiation are correct, they will give us more than just the correct statistical partition or distribution of the internal energy of the molecules. During absorption and emission of radiation there is also present a transfer of momentum to the molecules; this means that just the interaction of radiation and molecules leads to a velocity distribution of the latter. This must clearly be the same as the velocity distribution which molecules acquire as the result of their mutual interaction by collisions, that is, it must coincide with the Maxwell distribution. We must require that the mean kinetic energy which a molecule (per degree of freedom) acquires in a Planck radiation field of temperature T be
    \(\frac{k_B T}{2}\);
    this must be valid regardless of the nature of the molecules and independent of frequencies which the molecules absorb and emit. In this paper we wish to verify that this far-reaching requirement is, indeed, satisfied quite generally; as a result of this our simple hypotheses about the emission and absorption of radiation acquire new supports.

    In order to obtain this result, however, we must enlarge, in a definite way, the previous fundamental hypotheses which were related entirely to the exchange of energy. We are faced with this questions: Does the molecule suffer a push, when it absorbs or emits the energy \(\epsilon\)? As an example we consider, from the classical point of view, the emission of radiation. If a body emits the energy \(\epsilon\), it acquires a backward thrust [impulse] \(\frac{\epsilon}{c}\) if all the radiation \(\epsilon\) is radiated in the same direction. If, however, the radiation occurs through a spatially symmetric process, for example, spherical waves, there is no recoil at all. This alternative also plays a role in the quantum theory of radiation. If a molecule, in going from one possible quantum theoretic state to another, absorbs or elites the energy \(\epsilon\) in the form of radiation, such an elementary process can be looked upon a partly or fully directed in space, or also as a symmetric (non-directed) one. It turns out that we obtain a theory that is free of contradictions only if we consider the above elementary processes as being fully directed events; herein lies the principal result of the considerations that follow.
    http://astro1.panet.utoledo.edu/~ljc/einstein_ab.pdf (also see the Results section for the \(h \nu /c \) formulation.)

    That's the Dirac–Pauli representation of a single-particle, four-component wave equation of relativistic quantum mechanics, not the multi-particle Dirac field of Quantum Electrodynamics.

    https://arxiv.org/abs/hep-ph/9503416
    http://journals.aps.org/pr/abstract/10.1103/PhysRev.78.29 ( http://www.physics.drexel.edu/~bob/Quantum_Papers/Foldy-Wouthuysen.pdf )
     
    Last edited: Jan 7, 2017
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