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<p class="MsoNormal"><span style="font-size:14.0pt;color:#1F497D">Andre: <o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:14.0pt;color:#1F497D">Does your phrase “trispatial” represent the same and traditional 3D space, like “4D” represents the four dimensional “space-time” phrase?<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:14.0pt;color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-size:14.0pt;color:#1F497D">I personally think that the particles (“self-looped, in-phase oscillations”) should be modeled in the traditional 3D space.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:14.0pt;color:#1F497D">Chandra.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D"><o:p> </o:p></span></p>
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<p class="MsoNormal"><b><span style="font-size:11.0pt;font-family:"Calibri",sans-serif">From:</span></b><span style="font-size:11.0pt;font-family:"Calibri",sans-serif"> General [mailto:general-bounces+chandra.roychoudhuri=uconn.edu@lists.natureoflightandparticles.org]
<b>On Behalf Of </b>André Michaud<br>
<b>Sent:</b> Friday, January 12, 2018 11:58 PM<br>
<b>To:</b> general@lists.natureoflightandparticles.org<br>
<b>Subject:</b> Re: [General] closed paths<o:p></o:p></span></p>
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<p class="MsoNormal"><o:p> </o:p></p>
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<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;color:black">Hi Richard, Andrew and all,<br>
<br>
I agree Richard that the way you describe this seems to me the most accurate way that this could be described in 4D space geometry, particularly the apparent helicoidal circulation of the 2 charges of the carrying energy.<br>
<br>
I found that there is no way to represent the reciprocating motion of the photon "energy substance" in the frame of 4D space geometry due to the logical impossibility that during each transverse cycle, the growing magnetic aspect of this energy would have no
option but to occupy the same space as the closing in two charges by interpenetration, which is physically impossible if the "energy substance" has a physical existence, thus occupies a physical volume as it oscillates.<br>
<br>
This conundrum disappear in the trispatial geometry by the magnetic aspect growing and regressing in a different vectorial space.<br>
<br>
But this trispatial solution doesn't invalidate the 4D representtion of the helical model that you and others have come up with to deal with the twin charges. I can see how both representations directly correlate. I can even see your two fixed charges apparently
varying from max separation to zero and back at each cycle as they spiral along the way if I imagine myself following your photon on one side, looking at it.<br>
<br>
This is why I commented from the get go that your model seemed to me to be the best possible in a 4D space-time geometry.<br>
<br>
Best Regards<br>
--- André Michaud<br>
GSJournal admin<br>
<a href="http://www.gsjournal.net/">http://www.gsjournal.net/</a><br>
<a href="http://www.srpinc.org/">http://www.srpinc.org/</a> <br>
<br>
On Fri, 12 Jan 2018 20:15:58 -0800, Richard Gauthier wrote:<br>
<br>
Hello, <o:p></o:p></span></p>
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<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;color:black">Correction. This following sentence from my previous email derivation of Lambda-de-Broglie is INCORRECT:<o:p></o:p></span></p>
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<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;color:black">"The corresponding wavelength projected along the longitudinal axis of the photon-like object’s helical trajectory (whose forward angle theta is given by cos(theta) = v/c) turns out
to be lambda-de Broglie = lambda cos theta = h/(gamma mc) x cos(theta)= h/(gamma mc) x v/c = h/(gamma mv) which is the de Broglie wavelength."<o:p></o:p></span></p>
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<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;color:black"> <o:p></o:p></span></p>
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<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;color:black">The correct derivation of the de Broglie wavelength in my article takes the wave vector k=2pi/lambda of the helically circulating photon-like object and calculates K, the component
of this k vector along the helical axis by multiplying k by cos(theta) = v/c . This gives the K vector component along the helical axis, from which the de Broglie wavelength is derived from the formula K= 2pi /Lambda-de-Broglie i.e. Lambda-de-Broglie = 2pi
/ K which gives h/(gamma mv), the de Broglie wavelength.<o:p></o:p></span></p>
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<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;color:black">Thanks.<o:p></o:p></span></p>
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<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;color:black">Richard<o:p></o:p></span></p>
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<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;color:black"> <o:p>
</o:p></span></p>
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<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;color:black">On Jan 12, 2018, at 5:14 PM,
<a href="mailto:richgauthier@gmail.com">richgauthier@gmail.com</a> wrote:<o:p></o:p></span></p>
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<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;color:black"> <o:p>
</o:p></span></p>
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<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;color:black">Hello Andr</span><span style="font-size:10.5pt;font-family:"Calibri",sans-serif;color:black">é</span><span style="font-family:"Arial",sans-serif;color:black">and Andrew and all,
<o:p></o:p></span></p>
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<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;color:black"> <o:p>
</o:p></span></p>
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<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;color:black">De Broglie accepted that a relativistically-moving electron would have a higher internal frequency f given by hf=gamma mc^2. But he unfortunately didn’t associate that frequency f
with a circulating photon-like object composing the relativistically-moving electron. If you do you get a wavelength lambda (not the de Broglie wavelength) associated with that photon-like object (now called a spin-1/2 half photon) given by hf = hc/lambda
= gamma mc^2 . Solving for lambda of the photon-like object, you get lambda = h/gamma mc. If the photon-like object has a helical trajectory, then lambda is the wavelength of this photon-like object along its helical trajectory.
<b>The corresponding wavelength projected along the longitudinal axis of the photon-like object’s helical trajectory (whose forward angle theta is given by cos(theta) = v/c) turns out to be lambda-de Broglie = lambda cos theta = h/(gamma mc) x cos(theta)= h/(gamma
mc) x v/c = h/(gamma mv) which is the de Broglie wavelength.</b> This is explained in detail in my SPIE article “Electrons are spin-1/2 photons generating the de Broglie wavelength” (I now call them spin-1/2 half-photons) at<a href="https://richardgauthier.academia.edu/research#papers">https://richardgauthier.academia.edu/research#papers</a>(article
#17).<o:p></o:p></span></p>
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<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;color:black"> <o:p></o:p></span></p>
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<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;color:black">The circumference of the first Bohr orbit is one de Broglie wavelength long, as described by de Broglie. But the trajectory of the Bohr orbit electron (composed of a spin-1/2 charged
half-photon) would be a helical trajectory with many wavelengths lambda = h/(gamma mc) along this helical trajectory, whose closed helical axis length is one de Broglie wavelength. (The actual quantum mechanical description of the first electron orbital for
hydrogen is very different than the Bohr atom picture of course.)<o:p></o:p></span></p>
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<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;color:black"> <o:p></o:p></span></p>
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<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;color:black">Richard<o:p></o:p></span></p>
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<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;color:black"> <o:p>
</o:p></span></p>
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<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;color:black">On Jan 12, 2018, at 1:43 PM, André Michaud <<a href="mailto:srp2@srpinc.org">srp2@srpinc.org</a>> wrote:<o:p></o:p></span></p>
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<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;color:black"> <o:p>
</o:p></span></p>
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<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;color:black"> <o:p></o:p></span></p>
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<p style="mso-margin-top-alt:0in;margin-right:0in;margin-bottom:10.0pt;margin-left:0in">
<span style="font-family:"Calibri",sans-serif">Dear Andrew,</span><span style="font-family:"Arial",sans-serif"><o:p></o:p></span></p>
<p style="mso-margin-top-alt:0in;margin-right:0in;margin-bottom:10.0pt;margin-left:0in">
<span style="font-family:"Calibri",sans-serif">Indeed, this is a fundmental question.</span><span style="font-family:"Arial",sans-serif"><o:p></o:p></span></p>
<p style="mso-margin-top-alt:0in;margin-right:0in;margin-bottom:10.0pt;margin-left:0in">
<span style="font-family:"Calibri",sans-serif">I will try to explain how I see this. Note that my angle of "observation", so to speak, is strictly electromagnetic, meaning that to me energy, either free moving or stabilized as the invariant rest mass of elementary
particles such as the electron, can only be "electromagnetic" in nature.</span><span style="font-family:"Arial",sans-serif"><o:p></o:p></span></p>
<p style="mso-margin-top-alt:0in;margin-right:0in;margin-bottom:10.0pt;margin-left:0in">
<span style="font-family:"Calibri",sans-serif">We know that the electromagnetic oscillation is transverse with respect to the direction of motion. We know then that the "frequency" of a localized photon can only be the actual number of times the transverse
electromagnetic mutual induction cycle of its electric and magnetic aspects occur in 1 second.</span><span style="font-family:"Arial",sans-serif"><o:p></o:p></span></p>
<p style="mso-margin-top-alt:0in;margin-right:0in;margin-bottom:10.0pt;margin-left:0in">
<span style="font-family:"Calibri",sans-serif">BUT, the wavelength that we associate with it happens to be the "longitudinal distance" that the photon will travel during one of its cycles at velocity c</span><span style="font-family:"Arial",sans-serif"><o:p></o:p></span></p>
<p style="mso-margin-top-alt:0in;margin-right:0in;margin-bottom:10.0pt;margin-left:0in">
<span style="font-family:"Calibri",sans-serif">The deBroglie wavelenth is of the same nature. It is the longitudinal distance that the carrying-photon of the electron can travel during one of its cycle, which happens to be the length of the Bohr orbit because
it is now slowed down by having to "carry", so to speak, the innert mass of the electron.</span><span style="font-family:"Arial",sans-serif"><o:p></o:p></span></p>
<p style="mso-margin-top-alt:0in;margin-right:0in;margin-bottom:10.0pt;margin-left:0in">
<span style="font-family:"Calibri",sans-serif">Note: My blooper in my previous answer. the correct equations are of course: E= hf and E=(h c)/lambda.</span><span style="font-family:"Arial",sans-serif"><o:p></o:p></span></p>
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<span style="font-family:"Calibri",sans-serif">A free moving photon of 4.359743805E-18 J, has frequency of 6.579683917E15 Hz and a wavelength of 4.556335256E-8 m when calculated with lambda=hc</span><span lang="EN-CA" style="font-family:"Calibri",sans-serif">/E.</span><span style="font-family:"Arial",sans-serif"><o:p></o:p></span></p>
<p style="mso-margin-top-alt:0in;margin-right:0in;margin-bottom:10.0pt;margin-left:0in">
<span lang="EN-CA" style="font-family:"Calibri",sans-serif">A photon of 4.359743805E-18 j carrying an electron still has frequency of 6.579683917E15 Hz, but has an effective longitudinal wavelength of 3.32491846E-10 m only, that is the length of the Bohr orbit,
whether you calculate it with Lambda=h/mv or with lambda = hv/E, because when you equate (h/mv)=(hv/E) and simplify, you end up with E=mv^2, which restitutes 4.359743805E-18j, that is the amount of electromagnetic energy that carries the electron.</span><span style="font-family:"Arial",sans-serif"><o:p></o:p></span></p>
<p style="mso-margin-top-alt:0in;margin-right:0in;margin-bottom:10.0pt;margin-left:0in">
<span lang="EN-CA" style="font-family:"Calibri",sans-serif">My interpretation of the wavelength is that I don't think that anything is actually "longitudinally waving" with regard to the displacement of electromagnetic energy, whether free photons or electron
carrier-photon. From my perspective, the "classical wavelength" simply is synonymous with "distance travelled in space on a longitudinal trajectory during one electromagnetic cycle of the photon energy".</span><span style="font-family:"Arial",sans-serif"><o:p></o:p></span></p>
<p style="mso-margin-top-alt:0in;margin-right:0in;margin-bottom:10.0pt;margin-left:0in">
<span lang="EN-CA" style="font-family:"Calibri",sans-serif">The actual "waving" can only be cyclically transverse in my view and I found that the maximum amplitude of this waving for any photon corresponds to (lambda alpha)/(2 pi), so the longitudinal wavelength
is nevertheless related to the EM transverse frequency when related to alpha/(2 pi).
</span><span style="font-family:"Arial",sans-serif"><o:p></o:p></span></p>
<p style="mso-margin-top-alt:0in;margin-right:0in;margin-bottom:10.0pt;margin-left:0in">
<span style="font-family:"Calibri",sans-serif">I analyse this from another perspective in section "Defining a Distance Based Quantum of Action" in pages 8 and 9 of this paper:</span><span style="font-family:"Arial",sans-serif"><o:p></o:p></span></p>
<p style="mso-margin-top-alt:0in;margin-right:0in;margin-bottom:10.0pt;margin-left:0in">
<span style="font-family:"Calibri",sans-serif"><a href="https://www.omicsonline.org/open-access/the-last-challenge-of-modern-physics-2090-0902-1000217.pdf">https://www.omicsonline.org/open-access/the-last-challenge-of-modern-physics-2090-0902-1000217.pdf</a></span><span style="font-family:"Arial",sans-serif"><o:p></o:p></span></p>
<p style="mso-margin-top-alt:0in;margin-right:0in;margin-bottom:10.0pt;margin-left:0in">
<span style="font-family:"Calibri",sans-serif">I hope I am not confusing things still more with my unorthodox viewpoints.</span><span style="font-family:"Arial",sans-serif"><o:p></o:p></span></p>
<p style="mso-margin-top-alt:0in;margin-right:0in;margin-bottom:10.0pt;margin-left:0in">
<span style="font-family:"Calibri",sans-serif">Best Regards</span><span style="font-family:"Arial",sans-serif"><o:p></o:p></span></p>
<p style="mso-margin-top-alt:0in;margin-right:0in;margin-bottom:10.0pt;margin-left:0in">
<span style="font-family:"Calibri",sans-serif">André</span><span style="font-family:"Arial",sans-serif"><o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif"><br>
---<br>
André Michaud<br>
GSJournal admin<br>
<a href="http://www.gsjournal.net/">http://www.gsjournal.net/</a><br>
<a href="http://www.srpinc.org/">http://www.srpinc.org/</a><br>
<br>
On Fri, 12 Jan 2018 15:29:33 -0500, Andrew Meulenberg wrote: <o:p></o:p></span></p>
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<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">Dear André,<o:p></o:p></span></p>
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<p class="MsoNormal"><span style="font-family:"Calibri",sans-serif">Thanks for your comments; even tho they reflect our differing models.</span><span style="font-family:"Arial",sans-serif"><o:p></o:p></span></p>
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<p class="MsoNormal"><span style="font-family:"Calibri",sans-serif">A question, which answer was/is important to me, comes from your response to item 1. If "E=(lambda h)/c," then this refers only to light speed items (light) in a medium with refractive index
of 1. However, the de Broglie wavelength (lambda = h)/mv) only applies to bodies with mass. The de Broglie frequency is independent of medium and applies to massive items. So, I doubt that de Broglie would have equated the frequency and wavelength relations.</span><span style="font-family:"Arial",sans-serif"><o:p></o:p></span></p>
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<p class="MsoNormal"><span style="font-family:"Arial",sans-serif"> <o:p></o:p></span></p>
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<p class="MsoNormal"><span style="font-family:"Calibri",sans-serif">How do you interpret physic's emphasis on the wavelength and the ignoring of the frequency and what is actually waving? What is your guess/interpretation as to what is waving? Despite some
good descriptions and meaning of the deBroglie wavelength, I've not seen anyone in this group (or anywhere) give what I consider to be a valid answer to the frequency question, which I consider to be fundamental to the nature of the electron.</span><span style="font-family:"Arial",sans-serif"><o:p></o:p></span></p>
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<p class="MsoNormal"><span style="font-family:"Calibri",sans-serif">Andrew M.</span><span style="font-family:"Arial",sans-serif">
<o:p></o:p></span></p>
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</span><span style="font-family:"Calibri",sans-serif">_ _ _ </span><span style="font-family:"Arial",sans-serif"><o:p></o:p></span></p>
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<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">On Fri, Jan 12, 2018 at 1:15 PM, André Michaud <<a href="mailto:srp2@srpinc.org" target="_blank">srp2@srpinc.org</a>> wrote:
<o:p></o:p></span></p>
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<span style="font-family:"Calibri",sans-serif">Dear Andrew,</span><span style="font-family:"Arial",sans-serif"><o:p></o:p></span></p>
<p style="mso-margin-top-alt:0in;margin-right:0in;margin-bottom:10.0pt;margin-left:0in">
<span style="font-family:"Calibri",sans-serif">Thank you so much for your appreciation. I think no lead should be neglected in trying to figure out what is really happening at the fundamental level. I simply share those that I know of, when occasion arises.</span><span style="font-family:"Arial",sans-serif"><o:p></o:p></span></p>
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<span style="font-family:"Calibri",sans-serif">Relative to what you perceive as missing in relation to path independence, remember where de Broglie was at when he wrote this.
</span><span lang="EN-CA" style="font-family:"Calibri",sans-serif">This was 2 years before Schrödinger came up with the wave function. He was ears deep in the same sort of research that we are in now, about the same issues, but without the knowledge accumulated
since. This paper plus one other, I think, is what inspired Schrödinger to use the wave equation.</span><span style="font-family:"Arial",sans-serif"><o:p></o:p></span></p>
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<span style="font-family:"Calibri",sans-serif">To the 3 points you raised, here is what I think:</span><span style="font-family:"Arial",sans-serif"><o:p></o:p></span></p>
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<span style="font-family:"Calibri",sans-serif">1. I think that he saw frequency and wavelength as amounting to two equivalent references to the related amount of energy, and that he considered that mentioning one always implied the other. Lets remember that
E= hf, but that also E=(lambda h)/c.</span><span style="font-family:"Arial",sans-serif"><o:p></o:p></span></p>
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<span style="font-family:"Calibri",sans-serif">2. On page 509, when he writes : "the wave of frequency nu and of velocity c/beta must be in resonance over the whole length of the trajectory. This leads to condition", he was talking about the Bohr orbit in the
Bohr atom as a starting point, thus his reference to a "closed path" no doubt. This is how I interpret this.</span><span style="font-family:"Arial",sans-serif"><o:p></o:p></span></p>
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<span style="font-family:"Calibri",sans-serif">3. As for his use of the gamma factor, I have not specifically analyzed this particular issue, but I know now that he was deeply aware of Special relativity (thanks to Albrecht) and certainly was aware that the
energy level calculable for the Bohr orbit was sufficient to warrant a relativistic velocity of the electron on this orbit, if the electron actually ran this orbit (Heisenberg came to the conclusion in the same decade that it was possible that the electron
may not have been running this orbit, but could be stabilized at this distance without translating about the proton). Indeed, I also think that this is possible. What seems to matter is that in both cases, the energy level is the same. But yes, I also think
that the implications of his use of the gamma factor warrants investigation.</span><span style="font-family:"Arial",sans-serif"><o:p></o:p></span></p>
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<span style="font-family:"Calibri",sans-serif">For my mention of a "precision drift" of the velocity, I simply refer to the fact that if the electron were to orbit at Chip' inner radius limit distance, the more energetic electron's velocity would be higher,
while as the radius expands towards his outer radius limit, the less and less energetic electron's velocity would diminish in sync.</span><span style="font-family:"Arial",sans-serif"><o:p></o:p></span></p>
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<span style="font-family:"Calibri",sans-serif">The word precision, simply highlight that the velocity on the exact Bohr orbit is precise, while the possible spread of all orbits between r_outer and r_inner of Chip and Heisenberg equation amount to a precision
drift of this velocity.</span><span style="font-family:"Arial",sans-serif"><o:p></o:p></span></p>
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<span style="font-family:"Calibri",sans-serif">Hope this helps.</span><span style="font-family:"Arial",sans-serif"><o:p></o:p></span></p>
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<span style="font-family:"Calibri",sans-serif">Best Regards</span><span style="font-family:"Arial",sans-serif"><o:p></o:p></span></p>
<p style="mso-margin-top-alt:0in;margin-right:0in;margin-bottom:10.0pt;margin-left:0in">
<span style="font-family:"Calibri",sans-serif">André</span><span style="font-family:"Arial",sans-serif"><o:p></o:p></span></p>
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<span class="gmail-">---</span><br>
<span class="gmail-">André Michaud</span><br>
<span class="gmail-">GSJournal admin</span><br>
<a href="http://www.gsjournal.net/" target="_blank">http://www.gsjournal.net/</a><br>
<a href="http://www.srpinc.org/" target="_blank">http://www.srpinc.org/</a> <o:p>
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On Thu, 11 Jan 2018 14:39:08 -0500, Andrew Meulenberg wrote: <o:p></o:p></span></p>
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<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">Dear Andre,<o:p></o:p></span></p>
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<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">In your replies to Chip, you show what I consider to be a sign of the true scholar "the desire and ability to acquire, remember, and utilize information from many diverse sources." Thank you
for your link to de Broglie's 1923 paper. My French is not good enough to be sure that I was not interpreting his points as (rather than because of) his supporting of some of my views. In particular:
<o:p></o:p></span></p>
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<span style="font-family:"Arial",sans-serif">His emphasis on frequency (or period) rather than wavelength.<o:p></o:p></span></li><li class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto;mso-list:l0 level1 lfo1">
<span style="font-family:"Arial",sans-serif">His mention of closed path,<o:p></o:p></span></li><li class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto;mso-list:l0 level1 lfo1">
<span style="font-family:"Arial",sans-serif">I'm not quite sure what to do with his association of the relativistic gamma factor with the wave frequency. It looks interesting; but, I need to figure out the implications. Do you have an answer? It does get included
in his resonant energy relationship (which has a mv^2 rather than 1/2(mv^2) basis).<o:p></o:p></span></li></ol>
<p><span style="font-family:"Arial",sans-serif">His single mention of closed path, compared to his wave-based emphasis on frequency, misses statement of the importance of path independence of the closed or contour integral about non-singular regions. While
his wave functions provide cyclic examples of these closed paths, the importance to conservation laws is seldom (if ever?) mentioned in physics.<o:p></o:p></span></p>
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<p class="MsoNormal"><span style="font-family:"Arial",sans-serif"> <o:p></o:p></span></p>
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<p><span style="font-family:"Arial",sans-serif">I also have a question on a comment you made to Chip. In<o:p></o:p></span></p>
<p style="margin-left:30.0pt"><span style="font-family:"Arial",sans-serif">"</span>Heisenberg equation turns out to be de Broglie's equation for the Bohr orbit adapted to account for a<u> precision
</u>drift of the chosen velocity on either side of the selected velocity value on the ground orbital of the hydrogen atom."<span style="font-family:"Arial",sans-serif"><o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">you mention "</span>precision drift." Could you explain this a bit? I had originally thought that you meant
<i>precession drift</i>, which I consider to be very important (the basis of the de Broglie frequency). However, I don't think you had that in mind.<span style="font-family:"Arial",sans-serif"><o:p></o:p></span></p>
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<p class="MsoNormal">Andrew M.<span style="font-family:"Arial",sans-serif"> <o:p>
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<h3><span class="gmail-m6484677386908469894gmail-gd">André Michaud</span> <span class="gmail-m6484677386908469894gmail-go">
<<a href="mailto:srp2@srpinc.org" target="_blank">srp2@srpinc.org</a>></span><o:p></o:p></h3>
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<p class="MsoNormal"><span style="font-family:"Arial",sans-serif"> <o:p></o:p></span></p>
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<p class="MsoNormal"><span style="font-family:"Arial",sans-serif"> <o:p></o:p></span></p>
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<p class="MsoNormal"><span style="font-family:"Arial",sans-serif"> <o:p></o:p></span></p>
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<p style="mso-margin-top-alt:0in;margin-right:0in;margin-bottom:10.0pt;margin-left:0in">
Hi Chip,<span style="font-family:"Arial",sans-serif"><o:p></o:p></span></p>
<p style="mso-margin-top-alt:0in;margin-right:0in;margin-bottom:10.0pt;margin-left:0in">
As I signaled the typo in your equation on page 36, I forgot to mention something else that struck me (this is a part that I read carefully)<span style="font-family:"Arial",sans-serif"><o:p></o:p></span></p>
<p style="mso-margin-top-alt:0in;margin-right:0in;margin-bottom:10.0pt;margin-left:0in">
I notice that you mention that you noticed what you named a "beat frequency" with regard to the hydrogen ground state.<span style="font-family:"Arial",sans-serif"><o:p></o:p></span></p>
<p style="mso-margin-top-alt:0in;margin-right:0in;margin-bottom:10.0pt;margin-left:0in">
Just to mention that this the exact term that de Broglie used in French (un battement) to describe the resonance state that he associated with the hydrogen ground state. Here is a link to the paper that inspired Schrödinger to introduce the wave function on
account of this observation by de Broglie:<span style="font-family:"Arial",sans-serif"><o:p></o:p></span></p>
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<a href="http://www.academie-sciences.fr/pdf/dossiers/Broglie/Broglie_pdf/CR1923_p507.pdf" target="_blank">http://www.academie-sciences.fr/pdf/dossiers/Broglie/Broglie_pdf/CR1923_p507.pdf</a><span style="font-family:"Arial",sans-serif"><o:p></o:p></span></p>
<p style="mso-margin-top-alt:0in;margin-right:0in;margin-bottom:10.0pt;margin-left:0in">
The interesting part is in page 509.<span style="font-family:"Arial",sans-serif"><o:p></o:p></span></p>
<p style="mso-margin-top-alt:0in;margin-right:0in;margin-bottom:10.0pt;margin-left:0in">
I also noted that your outer and inner radii for the ground state can be directly related to Heisenberg's equation
<span style="font-family:"Arial",sans-serif"><o:p></o:p></span></p>
<p style="mso-margin-top-alt:0in;margin-right:0in;margin-bottom:10.0pt;margin-left:0in">
Heisenberg equation turns out to be de Broglie's equation for the Bohr orbit adapted to account for a precision drift of the chosen velocity on either side of the selected velocity value on the ground orbital of the hydrogen atom.<span style="font-family:"Arial",sans-serif"><o:p></o:p></span></p>
<p style="mso-margin-top-alt:0in;margin-right:0in;margin-bottom:10.0pt;margin-left:0in">
This range lies between your outer and inner radii.<br>
<br>
But you probably already were aware of this latter detail<span style="font-family:"Arial",sans-serif"><o:p></o:p></span></p>
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<span class="gmail-m6484677386908469894gmail-im">Best Regards</span><span style="font-family:"Arial",sans-serif"><o:p></o:p></span></p>
<p class="MsoNormal"><span class="gmail-m6484677386908469894gmail-im"><span style="font-family:"Arial",sans-serif">---</span></span><span style="font-family:"Arial",sans-serif"><br>
<span class="gmail-m6484677386908469894gmail-im">André Michaud</span><br>
<span class="gmail-m6484677386908469894gmail-im">GSJournal admin</span><o:p></o:p></span></p>
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