[General] research papers
Richard Gauthier
richgauthier at gmail.com
Wed Oct 21 15:32:05 PDT 2015
Hello Albert (and all),
I think your fundamental objection that you mentioned earlier can be answered below.
The left side of the big triangle in Figure 2 in my article is a purely mathematical unfolding of the path of the helical trajectory, to hopefully show more clearly the generation of de Broglie wavelengths from plane waves emitted by the actual charged photon moving along the helical trajectory. Nothing is actually moving off into space along this line.
Consider an electron moving with velocity v horizontally along the helical axis. Since in Figure 2 in my article, cos (theta) = v/c , the corresponding velocity of the charged photon along the helical path is v/ cos(theta) = c , the speed of the charged photon, which we knew already because the helical trajectory was defined so that this is the case. In a short time T, the electron has moved a distance Delectron = vT horizontally and the photon has moved a distance Dphoton = Delectron/cos(theta) =vT/cos(theta) = cT along its helical trajectory. A plane wave front emitted from the photon at the distance Dphoton = cT along the photon’s helical path will intersect the base of the big triangle (the helical axis) at the distance along the base given by Dwavefront = Dphoton / cos(theta) = cT/ (v/c) = T * (c^2)/v which means the intersection point of the plane wave with the helical axis is moving with a speed c^2/v which is the de Broglie wave’s phase velocity. The length of the de Broglie wave itself as shown previously from Figure 2 is Ldb = Lambda-photon / cos(theta) = h/(gamma mc) / (v/c) = h/(gamma mv). So as the electron moves with velocity v along the z-axis, de Broglie waves of length h/(gamma mv) produced along the z-axis are moving with velocity c^2/v along the z-axis. The de Broglie waves created by the circulating charged photon will speed away from the electron (but more will be produced) to take their place, one de Broglie wave during each period of the circulating charged photon (corresponding to the moving electron). As mentioned previously, the period of the circulating charged photon is 1/f = 1/(gamma mc^2/h) = h/(gamma mc^2/). As the electron speeds up (v and gamma increase) the de Broglie wavelengths h/(gamma mv) are shorter and move more slowly, following the speed formula c^2/v .
Unpublished graphic showing the generation of de Broglie waves from a moving charged photon along its helical trajectory. The corresponding moving electron is the red dot moving to the right on the red line. The charged photon is the blue dot moving at light speed along the helix.The blue dot has moves a distance of one charged photon wavelength h/(gamma mc) along the helix from the left corner of the diagram On the left diagonal line (representing the mathematically unrolled helix), the blue dots correspond to separations of 1 charged photon h/(gamma mc) wavelength along the helical axis. In this graphic, v/c = 0.5 so cos(theta)= 0.5 and theta= 60 degrees. The group velocity is c^2/v = c^2/0.5c = 2 c, the speed of the de Broglie waves along the horizontal axis . The distances between the intersection points on the horizontal line each correspond to 1 de Broglie wavelength, which in this example where v=0.5 c is h(gamma mv) = 2 x charged photon wavelength h/(gamma mc).
It is true that when the electron is at rest, the wave fronts emitted by the circulating charged photon all pass through the center of the circular path of the charged photon and do not intersect any helical axis, because no helical axis is defined for a resting electron, i.e. the pitch of the helix of the circulating charged photon is zero. For a very slowly moving electron, the pitch of the helix of the circulating charged photon is very small but non-zero, but the de Broglie wavelength is very large, much larger than the helical pitch. Perhaps you are confusing these two lengths — the helical pitch of the circulating charged photon and the de Broglie wavelength generated by the wave fronts emitted by the circulating charged photon. The pitch of the helix starts at zero (for v=0 of the electron) and reaches a maximum when the speed of the electron is c/sqrt(2) and theta = 45 degrees (see my charged photon paper) and then the helical pitch decreases towards zero as the speed of the electron further increases towards the speed of light. But the de Broglie wavelength Ldb starts very large (when the electron is moving very slowly) and decreases uniformly towards zero as the speed of the electron increases, as given by Ldb = h/gamma mv. It is the de Broglie wavelength generated by the charged photon that has predictive physical significance in diffraction and double-slit experiments while the helical pitch of the charged photon’s helical trajectory has no current predictive physical significance (though if experimental predictions based on the helical pitch could be made, this could be a test of the charged photon model).
I don’t have any comments yet on your concerns about the de Broglie wavelength that you just expressed to John W (below).
all the best,
Richard
> On Oct 21, 2015, at 12:42 PM, Dr. Albrecht Giese <genmail at a-giese.de> wrote:
>
> Dear John W and all,
>
> about the de Broglie wave:
>
> There are a lot of elegant derivations for the de Broglie wave length, that is true. Mathematical deductions. What is about the physics behind it?
>
> De Broglie derived this wave in his first paper in the intention to explain, why the internal frequency in a moving electron is dilated, but this frequency on the other hand has to be increased for an external observer to reflect the increase of energy. To get a result, he invented a "fictitious wave" which has the phase speed c/v, where v is the speed of the electron. And he takes care to synchronize this wave with the internal frequency of the electron. That works and can be used to describe the scattering of the electron at the double slit. - But is this physical understanding? De Broglie himself stated that this solution does not fulfil the expectation in a "complete theory". Are we any better today?
>
> Let us envision the following situation. An electron moves at moderate speed, say 0.1*c (=> gamma=1.02) . An observer moves parallel to the electron. What will the observer see or measure?
> The internal frequency of the electron will be observed by him as frequency = m0*c2/h , because in the observer's system the electron is at rest. The wave length of the wave leaving the electron (e.g. in the model of a circling photon) is now not exactly lambda1 = c/frequency , but a little bit larger as the rulers of the observer are a little bit contracted (by gamma = 1.02), so this is a small effect. What is now about the phase speed of the de Broglie wave? For an observer at rest it must be quite large as it is extended by the factor c/v which is 10. For the co-moving observer it is mathematically infinite (in fact he will see a constant phase). This is not explained by the time dilation (=2%), so not compatible. And what about the de Broglie wave length? For the co-moving observer, who is at rest in relation to the electron, it is lambdadB = h/(1*m*0), which is again infinite or at least extremely large. For the observer at rest there is lambdadB = h/(1.02*m*0.1c) . Also not comparable to the co-moving observer.
>
> To summarize: these differences are not explained by the normal SR effects. So, how to explain these incompatible results?
>
> Now let's assume, that the electron closes in to the double slit. Seen from the co-moving observer, the double slit arrangement moves towards him and the electron. What are now the parameters which will determine the scattering? The (infinite) de Broglie wave length? The phase speed which is 10*c ? Remember: For the co-moving observer the electron does not move. Only the double slit moves and the screen behind the double slit will be ca. 2% closer than in the standard case. But will that be a real change?
>
> I do not feel that this is a situation which in physically understood.
>
> Regards
> Albrecht
>
>
> Am 21.10.2015 um 16:34 schrieb John Williamson:
>> Dear all,
>>
>> The de Broglie wavelength is best understood, in my view, in one of two ways. Either read de Broglies thesis for his derivation (if you do not read french, Al has translated it and it is available online). Alternatively derive it yourself. All you need to do is consider the interference between a standing wave in one (proper frame) as it transforms to other relativistic frames. That is standing-wave light-in-a-box. This has been done by may folk, many times. Martin did it back in 1991. It is in our 1997 paper. One of the nicest illustrations I have seen is that of John M - circulated to all of you earlier in this series.
>>
>> It is real, and quite simple.
>>
>> Regards, John.
>> From: General [general-bounces+john.williamson=glasgow.ac.uk at lists.natureoflightandparticles.org <mailto:general-bounces+john.williamson=glasgow.ac.uk at lists.natureoflightandparticles.org>] on behalf of Dr. Albrecht Giese [genmail at a-giese.de <mailto:genmail at a-giese.de>]
>> Sent: Wednesday, October 21, 2015 3:14 PM
>> To: Richard Gauthier
>> Cc: Nature of Light and Particles - General Discussion; David Mathes
>> Subject: Re: [General] research papers
>>
>> Hello Richard,
>>
>> thanks for your detailed explanation. But I have a fundamental objection.
>>
>> Your figure 2 is unfortunately (but unavoidably) 2-dimensional, and that makes a difference to the reality as I understand it.
>>
>> In your model the charged electron moves on a helix around the axis of the electron (or equivalently the axis of the helix). That means that the electron has a constant distance to this axis. Correct? But in the view of your figure 2 the photon seems to start on the axis and moves away from it forever. In this latter case the wave front would behave as you write it.
>>
>> Now, in the case of a constant distance, the wave front as well intersects the axis, that is true. But this intersection point moves along the axis at the projected speed of the photon to this axis. - You can consider this also in another way. If the electron moves during a time, say T1, in the direction of the axis, then the photon will during this time T1 move a longer distance, as the length of the helical path (call it L) is of course longer than the length of the path of the electron during this time (call it Z). Now you will during the time T1 have a number of waves (call this N) on the helical path L. On the other hand, the number of waves on the length Z has also to be N. Because otherwise after an arbitrary time the whole situation would diverge. As now Z is smaller than L, the waves on the axis have to be shorter. So, not the de Broglie wave length. That is my understanding.
>>
>> In my present view, the de Broglie wave length has no immediate correspondence in the physical reality. I guess that the success of de Broglie in using this wave length may be understandable if we understand in more detail, what happens in the process of scattering of an electron at the double (or multiple) slits.
>>
>> Best wishes
>> Albrecht
>>
>>
>> Am 21.10.2015 um 06:28 schrieb
>> Richard Gauthier:
>>> Hello Albrecht,
>>>
>>> Thank you for your effort to understand the physical process described geometrically in my Figure 2. You have indeed misunderstood the Figure as you suspected. The LEFT upper side of the big 90-degree triangle is one wavelength h/(gamma mc) of the charged photon, mathematically unrolled from its two-turned helical shape (because of the double-loop model of the electron) so that its full length h/(gamma mc) along the helical trajectory can be easily visualized. The emitted wave fronts described in my article are perpendicular to this mathematically unrolled upper LEFT side of the triangle (because the plane waves emitted by the charged photon are directed along the direction of the helix when it is coiled (or mathematically uncoiled), and the plane wave fronts are perpendicular to this direction). The upper RIGHT side of the big 90-degree triangle corresponds to one of the plane wave fronts (of constant phase along the wave front) emitted at one wavelength lambda = h/(gamma mc) of the helically circulating charged photon. The length of the horizontal base of the big 90-degree triangle, defined by where this upper RIGHT side of the triangle (the generated plane wave front from the charged photon) intersects the horizontal axis of the helically-moving charged photon, is the de Broglie wavelength h/(gamma mv) of the electron model (labeled in the diagram). By geometry the length (the de Broglie wavelength) of this horizontal base of the big right triangle in the Figure is equal to the top left side of the triangle (the photon wavelength h/(gamma mc) divided (not multiplied) by cos(theta) = v/c because we are calculating the hypotenuse of the big right triangle starting from the upper LEFT side of this big right triangle, which is the adjacent side of the big right triangle making an angle theta with the hypotenuse.
>>>
>>> What you called the projection of the charged photon’s wavelength h/(gamma mc) onto the horizontal axis is actually just the distance D that the electron has moved with velocity v along the x-axis in one period T of the circulating charged photon. That period T equals 1/f = 1/(gamma mc^2/h) = h/(gamma mc^2). By the geometry in the Figure, that distance D is the adjacent side of the smaller 90-degree triangle in the left side of the Figure, making an angle theta with cT, the hypotenuse of that smaller triangle, and so D = cT cos (theta) = cT x v/c = vT , the distance the electron has moved to the right with velocity v in the time T. In that same time T one de Broglie wavelength has been generated along the horizontal axis of the circulating charged photon.
>>>
>>> I will answer your question about the double slit in a separate e-mail.
>>>
>>> all the best,
>>> Richard
>>>
>>>> On Oct 20, 2015, at 10:06 AM, Dr. Albrecht Giese <genmail at a-giese.de <mailto:genmail at a-giese.de>> wrote:
>>>>
>>>> Hello Richard,
>>>>
>>>> thank you for your explanations. I would like to ask further questions and will place them into the text below.
>>>>
>>>> Am 19.10.2015 um 20:08 schrieb Richard Gauthier:
>>>>> Hello Albrecht,
>>>>>
>>>>> Thank your for your detailed questions about my electron model, which I will answer as best as I can.
>>>>>
>>>>> My approach of using the formula e^i(k*r-wt) = e^i (k dot r minus omega t) for a plane wave emitted by charged photons is also used for example in the analysis of x-ray diffraction from crystals when you have many incoming parallel photons in free space moving in phase in a plane wave. Please see for example <http://www.pa.uky.edu/%7Ekwng/phy525/lec/lecture_2.pdf>http://www.pa.uky.edu/~kwng/phy525/lec/lecture_2.pdf <http://www.pa.uky.edu/~kwng/phy525/lec/lecture_2.pdf> . When Max Born studied electron scattering using quantum mechanics (where he used PHI*PHI of the quantum wave functions to predict the electron scattering amplitudes), he also described the incoming electrons as a plane wave moving forward with the de Broglie wavelength towards the target. I think this is the general analytical procedure used in scattering experiments. In my charged photon model the helically circulating charged photon, corresponding to a moving electron, is emitting a plane wave of wavelength lambda = h/(gamma mc) and frequency f=(gamma mc^2)/h along the direction of its helical trajectory, which makes a forward angle theta with the helical axis given by cos (theta)=v/c. Planes of constant phase emitted from the charged photon in this way intersect the helical axis of the charged photon. When a charged photon has traveled one relativistic wavelength lambda = h/(gamma mc) along the helical axis, the intersection point of this wave front with the helical axis has traveled (as seen from the geometry of Figure 2 in my charged photon article) a distance lambda/cos(theta) = lambda / (v/c) = h/(gamma mv) i.e the relativistic de Broglie wavelength along the helical axis.
>>>> Here I have a question with respect to your Figure 2. The circling charged photon is accompanied by a wave which moves at any moment in the direction of the photon on its helical path. This wave has its normal wavelength in the direction along this helical path. But if now this wave is projected onto the axis of the helix, which is the axis of the moving electron, then the projected wave will be shorter than the original one. So the equation will not be lambdadeBroglie = lambdaphoton / cos theta , but: lambdadeBroglie = lambdaphoton * cos theta . The result will not be the (extended) de Broglie wave but a shortened wave. Or do I completely misunderstand the situation here?
>>>>
>>>> Or let's use another view to the process. Lets imagine a scattering process of the electron at a double slit. This was the experiment where the de Broglie wavelength turned out to be helpful.
>>>> So, when now the electron, and that means the cycling photon, approaches the slits, it will approach at a slant angle theta at the layer which has the slits. Now assume the momentary phase such that the wave front reaches two slits at the same time (which means that the photon at this moment moves downwards or upwards, but else straight with respect to the azimuth). This situation is similar to the front wave of a single normal photon which moves upwards or downwards by an angle theta. There is now no phase difference between the right and the left slit. Now the question is whether this coming-down (or -up) will change the temporal sequence of the phases (say: of the maxima of the wave). This distance (by time or by length) determines at which angle the next interference maxima to the right or to the left will occur behind the slits.
>>>>
>>>> To my understanding the temporal distance will be the same distance as of wave maxima on the helical path of the photon, where the latter is lambda1 = c / frequency; frequency = (gamma*mc2) / h. So, the geometric distance of the wave maxima passing the slits is lambda1 = c*h / (gamma*mc2). Also here the result is a shortened wavelength rather than an extended one, so not the de Broglie wavelength.
>>>>
>>>> Again my question: What do I misunderstand?
>>>>
>>>> For the other topics of your answer I essentially agree, so I shall stop here.
>>>>
>>>> Best regards
>>>> Albrecht
>>>>
>>>>>
>>>>> Now as seen from this geometry, the slower the electron’s velocity v, the longer is the electron’s de Broglie wavelength — also as seen from the relativistic de Broglie wavelength formula Ldb = h/(gamma mv). For a resting electron (v=0) the de Broglie wavelength is undefined in this formula as also in my model for v = 0. Here, for stationary electron, the charged photon’s emitted wave fronts (for waves of wavelength equal to the Compton wavelength h/mc) intersect the axis of the circulating photon along its whole length rather than at a single point along the helical axis. This condition corresponds to the condition where de Broglie said (something like) that the electron oscillates with the frequency given by f = mc^2/h for the stationary electron, and that the phase of the wave of this oscillating electron is the same at all points in space. But when the electron is moving slowly, long de Broglie waves are formed along the axis of the moving electron.
>>>>>
>>>>> In this basic plane wave model there is no limitation on how far to the sides of the charged photon the plane wave fronts extend. In a more detailed model a finite side-spreading of the plane wave would correspond to a pulse of many forward moving electrons that is limited in both longitudinal and lateral extent (here a Fourier description of the wave front for a pulse of electrons of a particular spatial extent would probably come into play), which is beyond the present description.
>>>>>
>>>>> You asked what an observer standing beside the resting electron, but not in the plane of the charged photon's internal circular motion) would observe as the circulating charged photon emits a plane wave long its trajectory. The plane wave’s wavelength emitted by the circling charged photon would be the Compton wavelength h/mc. So when the charged photon is moving more towards (but an an angle to) the stationary observer, he would observe a wave of wavelength h/mc (which you call c/ny where ny is the frequency of charged photon’s orbital motion) coming towards and past him. This is not the de Broglie wavelength (which is undefined here and is only defined on the helical axis of the circulating photon for a moving electron) but is the Compton wavelength h/mc of the circulating photon of a resting electron. As the charged photon moves more away from the observer, he would observe a plane wave of wavelength h/mc moving away from him in the direction of the receding charged photon. But it is more complicated than this, because the observer at the side of the stationary electron (circulating charged photon) will also be receiving all the other plane waves with different phases emitted at other angles from the circulating charged photon during its whole circular trajectory. In fact all of these waves from the charged photon away from the circular axis or helical axis will interfere and may actually cancel out or partially cancel out (I don’t know), leaving a net result only along the axis of the electron, which if the electron is moving, corresponds to the de Broglie wavelength along this axis. This is hard to visualize in 3-D and this is why I think a 3-D computer graphic model of this plane-wave emitting process for a moving or stationary electron would be very helpful and informative.
>>>>>
>>>>> You asked about the electric charge of the charged photon and how it affects this process. Clearly the plane waves emitted by the circulating charged photon have to be different from the plane waves emitted by an uncharged photon, because these plane waves generate the quantum wave functions PHI that predict the probabilities of finding electrons or photons respectively in the future from their PHI*PHI functions. Plus the charged photon has to be emitting an additional electric field (not emitted by a regular uncharged photon), for example caused by virtual uncharged photons as described in QED, that produces the electrostatic field of a stationary electron or the electro-magnetic field around a moving electron.
>>>>>
>>>>> I hope this helps. Thanks again for your excellent questions.
>>>>>
>>>>> with best regards,
>>>>> Richard
>>>>>
>>>>>
>>>>>> On Oct 19, 2015, at 8:13 AM, Dr. Albrecht Giese < <mailto:genmail at a-giese.de>genmail at a-giese.de <mailto:genmail at a-giese.de>> wrote:
>>>>>>
>>>>>> Richard:
>>>>>>
>>>>>> I am still busy to understand the de Broglie wavelength from your model. I think that I understand your general idea, but I would like to also understand the details.
>>>>>>
>>>>>> If a photon moves straight in the free space, how does the wave look like? You say that the photon emits a plane wave. If the photon is alone and moves straight, then the wave goes with the photon. No problem. And the wave front is in the forward direction. Correct? How far to the sides is the wave extended? That may be important in case of the photon in the electron.
>>>>>>
>>>>>> With the following I refer to the figures 1 and 2 in your paper referred in your preceding mail.
>>>>>>
>>>>>> In the electron, the photon moves according to your model on a circuit. It moves on a helix when the electron is in motion. But let take us first the case of the electron at rest, so that the photon moves on this circuit. In any moment the plane wave accompanied with the photon will momentarily move in the tangential direction of the circuit. But the direction will permanently change to follow the path of the photon on the circuit. What is then about the motion of the wave? The front of the wave should follow this circuit. Would an observer next to the electron at rest (but not in the plane of the internal motion) notice the wave? This can only happen, I think, if the wave does not only propagate on a straight path forward but has an extension to the sides. Only if this is the case, there will be a wave along the axis of the electron. Now an observer next to the electron will see a modulated wave coming from the photon, which will be modulated with the frequency of the rotation, because the photon will in one moment be closer to the observer and in the next moment be farer from him. Which wavelength will be noticed by the observer? It should be lambda = c / ny, where c is the speed of the propagation and ny the frequency of the orbital motion. But this lambda is by my understanding not be the de Broglie wave length.
>>>>>>
>>>>>> For an electron at rest your model expects a wave with a momentarily similar phase for all points in space. How can this orbiting photon cause this? And else, if the electron is not at rest but moves at a very small speed, then the situation will not be very different from that of the electron at rest.
>>>>>>
>>>>>> Further: What is the influence of the charge in the photon? There should be a modulated electric field around the electron with a frequency ny which follows also from E = h*ny, with E the dynamical energy of the photon. Does this modulated field have any influence to how the electron interacts with others?
>>>>>>
>>>>>> Some questions, perhaps you can help me for a better understanding.
>>>>>>
>>>>>> With best regards and thanks in advance
>>>>>> Albrecht
>>>>>>
>>>>>> PS: I shall answer you mail from last night tomorrow.
>>>>>>
>>>>>>
>>>>>> Am 14.10.2015 um 22:32 schrieb Richard Gauthier:
>>>>>>> Hello Albrecht,
>>>>>>>
>>>>>>> I second David’s question. The last I heard authoritatively, from cosmologist Sean Carroll - "The Particle at the End of the Universe” (2012), is that fermions are not affected by the strong nuclear force. If they were, I think it would be common scientific knowledge by now.
>>>>>>>
>>>>>>> You wrote: "I see it as a valuable goal for the further development to find an answer (a physical answer!) to the question of the de Broglie wavelength."
>>>>>>> My spin 1/2 charged photon model DOES give a simple physical explanation for the origin of the de Broglie wavelength. The helically-circulating charged photon is proposed to emit a plane wave directed along its helical path based on its relativistic wavelength lambda = h/(gamma mc) and relativistic frequency f=(gamma mc^2)/h. The wave fronts of this plane wave intersect the axis of the charged photon’s helical trajectory, which is the path of the electron being modeled by the charged photon, creating a de Broglie wave pattern of wavelength h/(gamma mv) which travels along the charged photon’s helical axis at speed c^2/v. For a moving electron, the wave fronts emitted by the charged photon do not intersect the helical axis perpendicularly but at an angle (see Figure 2 of my SPIE paper at <https://www.academia.edu/15686831/Electrons_are_spin_1_2_charged_photons_generating_the_de_Broglie_wavelength>https://www.academia.edu/15686831/Electrons_are_spin_1_2_charged_photons_generating_the_de_Broglie_wavelength <https://www.academia.edu/15686831/Electrons_are_spin_1_2_charged_photons_generating_the_de_Broglie_wavelength> ) that is simply related to the speed of the electron being modeled. This physical origin of the electron’s de Broglie wave is similar to when a series of parallel and evenly-spaced ocean waves hits a straight beach at an angle greater than zero degrees to the beach — a wave pattern is produced at the beach that travels in one direction along the beach at a speed faster than the speed of the waves coming in from the ocean. But that beach wave pattern can't transmit “information” along the beach faster than the speed of the ocean waves, just as the de Broglie matter-wave can’t (according to special relativity) transmit information faster than light, as de Broglie recognized. As far as I know this geometric interpretation for the generation of the relativistic electron's de Broglie wavelength, phase velocity, and matter-wave equation is unique.
>>>>>>>
>>>>>>> For a resting (v=0) electron, the de Broglie wavelength lambda = h/(gamma mv) is not defined since one can’t divide by zero. It corresponds to the ocean wave fronts in the above example hitting the beach at a zero degree angle, where no velocity of the wave pattern along the beach can be defined.
>>>>>>>
>>>>>>> Schrödinger took de Broglie’s matter-wave and used it non-relativistically with a potential V to generate the Schrödinger equation and wave mechanics, which is mathematically identical in its predictions to Heisenberg’s matrix mechanics. Born interpreted Psi*Psi of the Schrödinger equation as the probability density for the result of an experimental measurement and this worked well for statistical predictions. Quantum mechanics was built on this de Broglie wave foundation and Born's probabilistic interpretation (using Hilbert space math.)
>>>>>>>
>>>>>>> The charged photon model of the electron might be used to derive the Schrödinger equation, considering the electron to be a circulating charged photon that generates the electron’s matter-wave, which depends on the electron’s variable kinetic energy in a potential field. This needs to be explored further, which I began in https://www.academia.edu/10235164/The_Charged-Photon_Model_of_the_Electron_Fits_the_Schrödinger_Equation <https://www.academia.edu/10235164/The_Charged-Photon_Model_of_the_Electron_Fits_the_Schr%C3%B6dinger_Equation> . Of course, to treat the electron relativistically requires the Dirac equation. But the spin 1/2 charged photon model of the relativistic electron has a number of features of the Dirac electron, by design.
>>>>>>>
>>>>>>> As to why the charged photon circulates helically rather than moving in a straight line (in the absence of diffraction, etc) like an uncharged photon, this could be the effect of the charged photon moving in the Higgs field, which turns a speed-of-light particle with electric charge into a less-than-speed-of-light particle with a rest mass, which in this case is the electron’s rest mass 0.511 MeV/c^2 (this value is not predicted by the Higgs field theory however.) So the electron’s inertia may also be caused by the Higgs field. I would not say that an unconfined photon has inertia, although it has energy and momentum but no rest mass, but opinions differ on this point. “Inertia” is a vague term and perhaps should be dropped— it literally means "inactive, unskilled”.
>>>>>>>
>>>>>>> You said that a faster-than-light phase wave can only be caused by a superposition of waves. I’m not sure this is correct, since in my charged photon model a single plane wave pattern emitted by the circulating charged photon generates the electron’s faster-than-light phase wave of speed c^2/v . A group velocity of an electron model may be generated by a superposition of waves to produce a wave packet whose group velocity equals the slower-than-light speed of an electron modeled by such an wave-packet approach.
>>>>>>>
>>>>>>> with best regards,
>>>>>>> Richard
>>>>>>>
>>
>>
>>
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