A Model for the Propagation of Visible Light and Other Rays

Discussion in 'Alternative Theories' started by yaldonTheory, Nov 8, 2016.

  1. exchemist Valued Senior Member

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    On further reflection, I think I would also take issue with the idea that electrons interact via an exchange of photons.

    In QFT the interaction is carried by virtual photons, so named because the are not photons, but disturbances in the EM field that can be modelled mathematically in a similar way to photons. More here: https://profmattstrassler.com/artic...ysics-basics/virtual-particles-what-are-they/
     
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  3. arfa brane call me arf Valued Senior Member

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    Well, physicists seem to make statements about particles and their interactions that "issues" can be taken with. You often see the statement "photons are massless", but that isn't accurate. Electrons do exchange real photons, but can also "exchange" virtual photons.

    Einstein's special theory first introduced (I think) the idea of relativistic mass, but nowadays this notion is considered problematic. Photons and electromagnetic radiation in general don't have mass in terms of kilograms, so what kind of mass do they have? How to explain accurately why the mass of a material body increases if its temperature does (because you "add" some heat energy)?

    I think there is a lot of confusing discussion about the consequences of special relativity, and what relativistic mass might mean, which is why a lot of physicists don't use it.
     
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  5. exchemist Valued Senior Member

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    When do electrons exchange real photons? I don't think I've ever heard of this. They emit and absorb photons under certain conditions, of course, but that is not "exchange". What do you have in mind?
     
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  7. arfa brane call me arf Valued Senior Member

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    A pair of electrons exchange a photon when one electron emits a photon and the other absorbs it. At least that's how I understand it. If that isn't an exchange then what is it?

    Also, the photon represents a real change in the phase of each electrons matter-field; that's something I understand as being true whether the photon is a real or a virtual one.
     
  8. exchemist Valued Senior Member

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    An exchange would, as the word implies, require a photon to go in BOTH directions, one from A to B and one from B to A.

    What you describe is a one-way process involving an emission, and then subsequently, after an arbitrary lapse of distance and time, an absorption by another electron that the photon may happen to encounter.
     
  9. NotEinstein Registered Senior Member

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    I remember seeing the term "exchange" in QM texts for a single (virtual) particle emission-and-absorption as well. I think this is (yet another) case of sloppy language usage in science.
     
  10. exchemist Valued Senior Member

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    What a shame. This is all a great pity, as it encourages the - seemingly widespread - confusion between photons and virtual photons, and between particles and the "virtual particles" of QFT more generally.

    This is a confusion that Strassler himself laments, in the link I provided. In fact he opens by saying we should forget the word "particle" in the phrase "virtual particle", as they are specifically NOT particles.

    I admit that I (being trained in chemistry) know almost no QFT, and in fact I came across the Strassler page looking for something to sort out my own confusion on the subject. I don't know how rigorous his explanation is, especially as it is not mathematical, but it seems a model of clarity to me.

    So now I am a fierce advocate of not muddling up the two - not least because, if one does muddle them up, it makes a dog's breakfast out of my own discipline!
     
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  11. arfa brane call me arf Valued Senior Member

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    Why? And why are photons and bosons in general called exchange particles by the physics community? Photons are also called force carriers and said to exchange the electromagnetic force between fermions.
    Yes, otherwise called an exchange between the two electrons which produces the EM force.
    And yet it's the language used by Feynman, and nowadays widely used. It's quite common to see this terminology in the literature. Have you tried googling "exchange particle" lately?

    And about the exchange of virtual photons: two electrons "bounce off" each other because of the Coulomb potential but exchange a virtual photon. This is what a Feynman diagram says if the photon (the wavy thing) has no temporal existence (alternately no spatial existence), i.e. is parallel to the space (resp. time) axis in the diagram. If the exchange particle is not parallel to either axis then it has both a spatial and temporal existence, i.e. represents a real photon.
     
    Last edited: Oct 6, 2017
  12. exchemist Valued Senior Member

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    Just read the link I posted, OK?
     
  13. arfa brane call me arf Valued Senior Member

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    I have. Could you just google "exchange particle" and at least see how many hits you get? Maybe read one of the links too?
     
  14. NotEinstein Registered Senior Member

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    I just did; it's used everywhere, I see. But that doesn't mean it isn't confusing (for laymen). The word "exchange" (in its general definition) suggests a back-and-forth, and a single (virtual) photon going from one electron to another isn't that. But alas, it's indeed standard terminology (or "jargon", if you will) now.
     
  15. arfa brane call me arf Valued Senior Member

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    The confusing thing about using the term "particle" is that quantum particles aren't really like particles at all, but the terminology has stuck. Even the link posted by exchemist has some confusing stuff in it.

    Perhaps that just underlines that trying to describe QFT in words (even with accompanying diagrams), can't really get close to how the mathematics does it; Feynman diagrams are just heuristics for the equations, and should not be interpreted as representing "particles" at all, is how I see it.
     
  16. NotEinstein Registered Senior Member

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    This suddenly reminded of the "wave packets" in QFT. You are absolutely right: "particle" is indeed a hopelessly confusing term in QFT contexts.

    That's my opinion as well: words are good and all, but when you really want to know QFT, die-hard maths is the only option.
     
  17. exchemist Valued Senior Member

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    There may be a partial resolution of this argument at the foot of the Wik entry here: https://en.wikipedia.org/wiki/Virtual_particle

    This makes the same distinctions that Strassler does. However, right at the bottom it indicates that there is a link between real and virtual particles by way of lifetime. It says in effect that a real particle (with a well-defined energy) can be seen as the long-lifetime special case of the virtual particle.
     
  18. el es Registered Senior Member

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    My layman understanding is that free electrons can not absorb or emit real photons, but they can exchange virtual photons resulting in repulsion.
     
  19. arfa brane call me arf Valued Senior Member

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    It depends how fast the electrons are moving and whether they then encounter a magnetic field, which gives them transverse acceleration. This is in fact the basis of synchrotron radiation, and the free electron laser.

    Actually that's a bit inaccurate, electrons emit radiation when they accelerate; I guess then it depends what you think a free electron is. Is it an electron which isn't affected by electric or magnetic fields? Or is it an electron in a vacuum, rather than in some medium like a metal?
     
    Last edited: Oct 10, 2017
  20. el es Registered Senior Member

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    263
    So, I should have specified that the free electrons be in the vacuum having no relative motion in an inertial reference frame.
    Would they then exchange real photons?
    Would they then exchange virtual photons if in proximity?
     
  21. yaldonTheory Registered Member

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    Obviously you have not read our work, due to the fact that we use Calculus throughout The Yaldon Particle Theory.

    When a mass (m) is under a net force that is equal to zero, then the energy will be potential energy.
    For example: When a mass is elevated and kept at a height of y meter(s), it will have potential energy \((E_p=mgy)\). In the same way, when a mass travels in empty space with a constant velocity, it will have a net force that is equal to zero (Newton's First Law). Then this mass will have potential energy that is equal to \(E_p=mv^2_c\); where \(v_c\) is the constant velocity. When this mass (m) has a variable velocity, then it will have kinetic energy.
    We will translate the above statement to Math using Newton's Laws:
    For Kinetic Energy:
    \(F(v)=\frac{m}{t}v\)
    \([F(v)=\frac{m}{t}v]\cdot dv\)
    \(\int\limits_0^{v_{max}}F(v)\cdot dv=\frac{m}{t}\int\limits_0^{v_{max}}v\cdot dv\)
    \(Fv_{max}=\frac{1}{2}\frac{m}{t}v^2_{max}\)
    \(F[v_{max}\cdot t]=\frac{1}{2}mv^2_{max}\)
    Where \(F\cdot s_{max}=E_k\)
    \(E_k=\frac{1}{2}mv^2_{max}\)
    For Potential Energy:
    \(F(s)=m\frac{v}{t}\)
    Where \(m\frac{v}{t}\) is a constant value; then the velocity will also be a constant \((v_c)\).
    After multiplying both sides of the above equation by ds and taking the limit from zero to \(s_{max}\):
    \(F(s)\int\limits_0^{s_{max}}ds=m\frac{v_c}{t}\int\limits_0^{s_{max}}ds\)
    \(F\cdot s_{max}=m\frac{v_c}{t}\cdot s_{max}\)
    Where \(F\cdot s_{max}=E_p\)
    \(E_p=mv_c\frac{s_{max}}{t}\)
    Where \(\frac{s_{max}}{t}=v_c\), since both are properties of the same mass (m)
    \(E_p=mv^2_c\)
    When \(E_k=E_p\) for the same mass (m)
    \(\frac{1}{2}mv^2_{max}=mv^2_c\)
    Then:
    \(\frac{1}{2}v^2_{max}=v^2_c\)​
    Or:
    \(v^2_{max}=2v^2_c\)​

    The above mathematical proof is not needed to explain The Yaldon Particle Theory, we just want to show the relation between the Laws of Newton and energy. Since the formulas \(E_k=\frac{1}{2}mv^2_{max}\) and \(E_p=mgy\) are accepted, then one must accept the Laws of Newton; due to the fact that these formulas have been derived from Newton's Laws. As one can see, we don't need a "rest mass" versus "relative mass" philosophy.
    As one can see, when a mass moves with a constant velocity; it will have potential energy. The Yaldon Particle Theory has arrived at this conclusion; due to the correct interpretation of Newton's Laws.
     
    Last edited: Oct 10, 2017
  22. arfa brane call me arf Valued Senior Member

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    One cannot see that though. Potential energy is position-dependent, as you say here: "For example: When a mass is elevated and kept at a height of y meter(s), it will have potential energy . . .".

    The height of y metres is a definition of a position. Potential energy depends on position, not constant velocity. Note that to define a position you also have to choose somewhere the position = 0 (in the case of height, 0 is ground level).
     
    Last edited: Oct 10, 2017
  23. yaldonTheory Registered Member

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    53
    In regard to time:
    Time is a property of a moving mass in a bounded system. Then a mass has to be in motion in order to have a value for time. As a mass moves with a velocity of v, it will have a certain value for the displacement (d).
    Then time (t) is merely the displacement over the velocity:
    \(t=\frac{d}{v}\)​
    If the mass is not in motion within a bounded system, then there will be no time. The Yaldon Particle Theory requires a mass in motion in a bounded system so that time can have a value.
    According to Newton's Second Law: \(F=m\frac{v}{t}\)
    When t=o, then v must also equal zero according to The Yaldon Particle Theory. Then when t=0 the formula will be \(F=m\frac{0}{0}\). Then the value for the force will not be infinity; it will be indeterminate. Newton's Second Law is not intended for a stationary mass. According to The Yaldon Particle Theory time will always be a property of a mass in motion within a bounded system.

    We created The Yaldon Particle Theory since we've noticed that current theories are using a lot of jargon and terminology which are in conflict with the established Laws of Newton. The Yaldon Particle Theory has discovered that all of these jargons and terminologies are not needed in order to explain phenomena. All that is needed in The Yaldon Particle Theory are the four basic assumed properties of a yaldon particle, and no other theory has these assumptions. These four assumed properties are an: average radius, average speed, mass, and they are perfectly elastic with no spin, charge, friction, nor any other fields surrounding them. With these four assumed properties, The Yaldon Particle Theory has been able to create a far more accurate model for the atom; with a more evenly distributed mass within the atom, and no need for "sub-shells" or electrons.
    Yaldon Theory's intent is to make Physics easier to understand and grasp, without needing theories with a lot of extra terminology that cannot coincide with the established Laws of Newton or one another.

    Once again, let us keep the topic of discussion focused to The Yaldon Particle Theory.
    Here is a link to the .pdf file for The Yaldon Particle Theory: The Explanation of Thermodynamics, Propagation of Rays, Related Phenomena, and the Model of the Atom.
    https://drive.google.com/file/d/0B-vl7kjDEZ8HV2lZN1dId3JpcUU/preview
     

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