<html><head><meta http-equiv="Content-Type" content="text/html charset=utf-8"></head><body style="word-wrap: break-word; -webkit-nbsp-mode: space; -webkit-line-break: after-white-space;" class=""><div class="">Hi Chip,</div><div class=""> Here you’re just objectively mistaken. The measurable momentum of a relativistic electron is p=gamma mv. Please look it up. The quantity p=E/c = gamma mc^2 / c = gamma mc is always greater than p = gamma mv .</div><div class=""> Richard</div><div class=""><br class=""></div><br class=""><div><blockquote type="cite" class=""><div class="">On Jan 25, 2017, at 6:02 AM, Chip Akins <<a href="mailto:chipakins@gmail.com" class="">chipakins@gmail.com</a>> wrote:</div><br class="Apple-interchange-newline"><div class=""><div class="WordSection1" style="page: WordSection1; font-family: Helvetica; font-size: 12px; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: normal; line-height: normal; orphans: auto; text-align: start; text-indent: 0px; text-transform: none; white-space: normal; widows: auto; word-spacing: 0px; -webkit-text-stroke-width: 0px;"><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""><span style="" class="">Hi Richard<o:p class=""></o:p></span></div><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""><span style="" class=""> </span></div><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""><span style="" class="">I do not feel that p=E/c is the total momentum, but rather the normal measureable momentum. Since it is the momentum measured longitudinally. It is then just one vector component of the total momentum, the total momentum is more likely pt = sqrt(2) E/c.<o:p class=""></o:p></span></div><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""><span style="" class=""> </span></div><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""><span style="" class="">I do not think there is a photon inside the electron. I think the energy in the electron can become a photon. That energy can be an electron, or a photon, but not both at the same time.<o:p class=""></o:p></span></div><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""><span style="" class=""> </span></div><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""><span style="" class="">Chip<o:p class=""></o:p></span></div><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""><span style="" class=""> </span></div><div class=""><div style="border-style: solid none none; border-top-color: rgb(225, 225, 225); border-top-width: 1pt; padding: 3pt 0in 0in;" class=""><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""><b class=""><span style="font-size: 11pt; font-family: Calibri, sans-serif;" class="">From:</span></b><span style="font-size: 11pt; font-family: Calibri, sans-serif;" class=""><span class="Apple-converted-space"> </span>General [<a href="mailto:general-bounces+chipakins=gmail.com@lists.natureoflightandparticles.org" style="color: purple; text-decoration: underline;" class="">mailto:general-bounces+chipakins=gmail.com@lists.natureoflightandparticles.org</a>]<span class="Apple-converted-space"> </span><b class="">On Behalf Of<span class="Apple-converted-space"> </span></b>Richard Gauthier<br class=""><b class="">Sent:</b><span class="Apple-converted-space"> </span>Wednesday, January 25, 2017 12:37 AM<br class=""><b class="">To:</b><span class="Apple-converted-space"> </span>Nature of Light and Particles - General Discussion <<a href="mailto:general@lists.natureoflightandparticles.org" style="color: purple; text-decoration: underline;" class="">general@lists.natureoflightandparticles.org</a>><br class=""><b class="">Subject:</b><span class="Apple-converted-space"> </span>Re: [General] light and particles group<o:p class=""></o:p></span></div></div></div><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""><o:p class=""> </o:p></div><div class=""><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class="">Hi Chip (and all), <o:p class=""></o:p></div></div><div class=""><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""> Thanks for summarizing your approach to calculating electron spin from your model.<o:p class=""></o:p></div></div><div class=""><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class="">You quoted Wikipedia:<o:p class=""></o:p></div></div><div class=""><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""><o:p class=""> </o:p></div></div><div class=""><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""> <span style="background-color: white; background-position: initial initial; background-repeat: initial initial;" class="">Please see PHOTON in Wikipedia… </span><i class="">“The photon also carries a quantity called spin angular momentum that does not depend on its frequency. The magnitude of its spin is sqrt(2) hbar and the component measured along its direction of motion, its helicity, must be ±ħ.”</i><o:p class=""></o:p></div></div><div class=""><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""><o:p class=""> </o:p></div></div><div class=""><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class="">I did vector calculations for the total angular momentum L= R x p of both my spin 1 hbar and spin 1/2 hbar transluminal energy quantum photon models. The results were as follows for L^2 , which is the square of the total angular momentum of the photon models.<o:p class=""></o:p></div></div><div class=""><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""><o:p class=""> </o:p></div></div><div class=""><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class="">For the spin 1 hbar photon model: L^2 = 2 hbar^2 + (Et)^2 where E is the energy of the photon and t is the time (arbitrary starting time) for the mathematical helical motion of the energy quantum.<o:p class=""></o:p></div></div><div class=""><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""><o:p class=""> </o:p></div></div><div class=""><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class="">For the spin 1/2 hbar charged photon model: L^2 = 2 (hbar/2)^2 + (Et)^2 for E and t as above.<o:p class=""></o:p></div></div><div class=""><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""><o:p class=""> </o:p></div></div><div class=""><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class="">This shows that as t -> 0 (remembering that the starting point of t is arbitrary in the models, the above total angular momentum equations for the photon models reduce to L= sqrt(2) hbar for the spin 1 photon model and L= sqrt(2) (hbar/2) for the spin 1/2 photon model. This is the quantum mechanical result for the total spin of a spin 1 photon (as mentioned above in the Wikipedia quote) , remembering that Sz = hbar and hbar/2 are the z-components of spin of a photon or electron respectively that are actually measured in photon and electron experiments. I had never previously done these calculations for the total spin of my spin 1 and spin 1/2 charged photon models (whose equations are summarized in my recent article at<span class="Apple-converted-space"> </span><a href="https://www.academia.edu/30899196/Transluminal_Energy_Quantum_Model_of_a_Spin-_Charged_Photon_Composing_an_Electron" style="color: purple; text-decoration: underline;" class="">https://www.academia.edu/30899196/Transluminal_Energy_Quantum_Model_of_a_Spin-_Charged_Photon_Composing_an_Electron</a> ) which is attached below. So the fact that the result for the total spin sqrt(2) hbar of my spin 1 model photon is consistent with the quantum mechanical result for a spin 1 photon is a very nice result. The expression (Et)^2 in the total angular momentum L^2 result is curious. But I think this part of the result is because in the calculation of L=R x p , R is measured from the origin (x=0, y=0, z=0) of the coordinate system in which the helical motion is given. Since for large values of t, R increases with time approximately as R=ct, while p = (E/c) sqrt(2) for the transluminal energy quantum of the photon models, the vector calculation of L= R x p =~ ct x (E/c) (sqrt (2)= Et is also increasing with time (which accounts for the (Et)^2 contribution to the total value of L^2 for both the spin 1 and spin 1/2 transluminal energy quantum photon models when t is large compared to the period t = 1/f of the photon.<o:p class=""></o:p></div></div><div class=""><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""><o:p class=""> </o:p></div></div><div class=""><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class="">Chip, I also noticed in your calculation for the spin of your electron model in your email below, you calculate the electron spin as s=r x p where p =E/c = gamma mc is the momentum of the helically moving photon. But we have seen that the transverse component of this total momentum p=E/c of the helically moving photon is only Ptrans = Eo/c = mc and not E/c = gamma mc. So when this Ptrans=Eo/c is multiplied by <span style="background-color: white; background-position: initial initial; background-repeat: initial initial;" class="">r = (c h)/(4pi E)=(c h)/(4pi gamma Eo) </span> in your model, the result is S = r x Ptrans = h/(4 pi gamma) = hbar/(2 gamma) and not hbar/2 as you state below in your email. And you have apparently left out the contribution of the spin of the helically moving photon itself in your calculation of the total spin of the helically moving photon model of the electron.<o:p class=""></o:p></div></div><div class=""><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""><o:p class=""> </o:p></div></div><div class=""><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""> I hope this is useful.<o:p class=""></o:p></div></div><div class=""><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""> Richard<o:p class=""></o:p></div></div><div class=""><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""><o:p class=""> </o:p></div></div><div class=""><div style="margin: 0in 0in 0.0001pt; font-size: 12pt; font-family: 'Times New Roman', serif;" class=""><o:p class=""> </o:p></div></div><div class=""><div style="margin: 0in 0in 0.0001pt; 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