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</o:shapelayout></xml><![endif]--></head><body bgcolor=white lang=EN-US link=blue vlink=purple><div class=WordSection1><p class=MsoNormal><span style='color:black'>Hi Grahame and Richard<o:p></o:p></span></p><p class=MsoNormal><span style='color:black'><o:p> </o:p></span></p><p class=MsoNormal><span style='color:black'>So what I am working to sort out is the correct view of the momentum vectors within the electron. Of course SR postulates that nothing within the electron can be moving faster than c.<o:p></o:p></span></p><p class=MsoNormal><span style='color:black'>I am working to determine if that is precisely correct.<o:p></o:p></span></p><p class=MsoNormal><span style='color:black'><o:p> </o:p></span></p><p class=MsoNormal><span style='color:black'>Here is a quote from Wikipedia regarding the spin of the photon ¡°The magnitude of the photon¡¯s spin is </span><!--[if gte msEquation 12]><m:oMath><m:rad><m:radPr><m:degHide m:val="on"/><span style='font-family:"Cambria Math",serif;color:black'><m:ctrlPr></m:ctrlPr></span></m:radPr><m:deg></m:deg><m:e><span style='font-family:"Cambria Math",serif;color:black'><m:r><m:rPr><m:scr m:val="roman"/><m:sty m:val="p"/></m:rPr>2</m:r></span></m:e></m:rad><span style='font-family:"Cambria Math",serif;color:black'><m:r><m:rPr><m:scr m:val="roman"/><m:sty m:val="p"/></m:rPr> </m:r><m:r><i>c</i></m:r></span></m:oMath><![endif]--><![if !msEquation]><span style='font-size:12.0pt;font-family:"Times New Roman",serif;position:relative;top:3.0pt;mso-text-raise:-3.0pt;mso-fareast-language:EN-US'><img width=31 height=21 id="_x0000_i1025" src="cid:image002.png@01D1D82B.3CA0F5B0"></span><![endif]><span style='color:black'> and the component measured along its direction of motion¡¦ must be ¡¾©¤.¡± This suggests that the wavefront in the photon is internal to the photon moving at</span><!--[if gte msEquation 12]><m:oMath><m:rad><m:radPr><m:degHide m:val="on"/><span style='font-family:"Cambria Math",serif;color:black'><m:ctrlPr></m:ctrlPr></span></m:radPr><m:deg></m:deg><m:e><span style='font-family:"Cambria Math",serif;color:black'><m:r><m:rPr><m:scr m:val="roman"/><m:sty m:val="p"/></m:rPr>2</m:r></span></m:e></m:rad><span style='font-family:"Cambria Math",serif;color:black'><m:r><m:rPr><m:scr m:val="roman"/><m:sty m:val="p"/></m:rPr> </m:r><m:r><i>c</i></m:r></span></m:oMath><![endif]--><![if !msEquation]><span style='font-size:12.0pt;font-family:"Times New Roman",serif;position:relative;top:3.0pt;mso-text-raise:-3.0pt;mso-fareast-language:EN-US'><img width=31 height=21 id="_x0000_i1025" src="cid:image002.png@01D1D82B.3CA0F5B0"></span><![endif]><span style='color:black'>.<o:p></o:p></span></p><p class=MsoNormal><span style='color:black'><o:p> </o:p></span></p><p class=MsoNormal><span style='color:black'>So if we visit again the equation for the momentum of a wave we can see that perhaps p=E/c has been simplified so much that part of the information is absent. We can see that the term should be p = (E/c^2)v, and perhaps that v in the electron can be </span><!--[if gte msEquation 12]><m:oMath><m:rad><m:radPr><m:degHide m:val="on"/><span style='font-family:"Cambria Math",serif;color:black'><m:ctrlPr></m:ctrlPr></span></m:radPr><m:deg></m:deg><m:e><span style='font-family:"Cambria Math",serif;color:black'><m:r><m:rPr><m:scr m:val="roman"/><m:sty m:val="p"/></m:rPr>2</m:r></span></m:e></m:rad><span style='font-family:"Cambria Math",serif;color:black'><m:r><m:rPr><m:scr m:val="roman"/><m:sty m:val="p"/></m:rPr> </m:r><m:r><i>c</i></m:r></span></m:oMath><![endif]--><![if !msEquation]><span style='font-size:12.0pt;font-family:"Times New Roman",serif;position:relative;top:3.0pt;mso-text-raise:-3.0pt;mso-fareast-language:EN-US'><img width=31 height=21 id="_x0000_i1025" src="cid:image002.png@01D1D82B.3CA0F5B0"></span><![endif]><span style='color:black'> as in the photon.<o:p></o:p></span></p><p class=MsoNormal><span style='color:black'><o:p> </o:p></span></p><p class=MsoNormal><span style='color:black'>But this complicates the issue of fully understanding the spin of the electron, and can cause us to begin to question exactly what it is we are measuring when we measure the electron¡¯s spin.<o:p></o:p></span></p><p class=MsoNormal><span style='color:black'><o:p> </o:p></span></p><p class=MsoNormal><span style='color:black'>Chip<o:p></o:p></span></p><p class=MsoNormal><span style='color:black'><o:p> </o:p></span></p><div><div style='border:none;border-top:solid #E1E1E1 1.0pt;padding:3.0pt 0in 0in 0in'><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+chipakins=gmail.com@lists.natureoflightandparticles.org] <b>On Behalf Of </b>Dr Grahame Blackwell<br><b>Sent:</b> Thursday, July 07, 2016 3:00 AM<br><b>To:</b> Nature of Light and Particles - General Discussion <general@lists.natureoflightandparticles.org><br><b>Subject:</b> Re: [General] double photon cycle, subjective v objective realities<o:p></o:p></span></p></div></div><p class=MsoNormal><o:p> </o:p></p><div><p class=MsoNormal><span style='font-size:10.0pt;font-family:"Arial",sans-serif;color:navy'>Thanks Richard,</span><o:p></o:p></p></div><div><p class=MsoNormal> <o:p></o:p></p></div><div><p class=MsoNormal><span style='font-size:10.0pt;font-family:"Arial",sans-serif;color:navy'>That's precisely what I've been trying to say, without in any way resting on any generally-accepted results that might be regarded as consequences of SR (and so open to question).</span><o:p></o:p></p></div><div><p class=MsoNormal> <o:p></o:p></p></div><div><p class=MsoNormal><span style='font-size:10.0pt;font-family:"Arial",sans-serif;color:navy'>If we agree that the transverse momentum component of the electron is a direct consequence of the rotational component of its formative photon (as I hope we do!) then that rotational component is acting at radius R of the electron at that time from its centre. Angular momentum is given by linear tangential momentum multiplied by radius - so angular momentum of the electron is mcR. Since mc is constant, R must also be constant if angular momentum is invariant (which I believe we agree it is).</span><o:p></o:p></p></div><div><p class=MsoNormal> <o:p></o:p></p></div><div><p class=MsoNormal><span style='font-size:10.0pt;font-family:"Arial",sans-serif;color:navy'>Just one further point: Richard, you refer to m as the electron's invariant mass. If we regard mass as that quality of an object that resists acceleration (and so is proportional to the instantaneous force required to induce an instantaneous acceleration), then my research indicates that the mass is <em><span style='font-family:"Arial",sans-serif'>not</span></em> invariant - though it will appear so from measurements taken within the electron's moving frame. My analysis shows that objective mass varies with speed and the relationship E = mc^2 is applicable only for an objectively static object/particle. The m referred to above, as I see it, is the objective rest-mass of the electron (i.e. its mass when objectively static), which corresponds to the energy required to maintain the formative structure of the electron (as opposed to that required to maintain its linear motion). This is of course constant.</span><o:p></o:p></p></div><div><p class=MsoNormal> <o:p></o:p></p></div><div><p class=MsoNormal><span style='font-size:10.0pt;font-family:"Arial",sans-serif;color:navy'>Best regards,</span><o:p></o:p></p></div><div><p class=MsoNormal><span style='font-size:10.0pt;font-family:"Arial",sans-serif;color:navy'>Grahame</span><o:p></o:p></p></div><blockquote style='border:none;border-left:solid navy 1.5pt;padding:0in 0in 0in 4.0pt;margin-left:3.75pt;margin-top:5.0pt;margin-right:0in;margin-bottom:5.0pt'><div><p class=MsoNormal><span style='font-size:10.0pt;font-family:"Arial",sans-serif'>----- Original Message ----- <o:p></o:p></span></p></div><div><p class=MsoNormal style='background:#E4E4E4'><b><span style='font-size:10.0pt;font-family:"Arial",sans-serif'>From:</span></b><span style='font-size:10.0pt;font-family:"Arial",sans-serif'> <a href="mailto:richgauthier@gmail.com" title="richgauthier@gmail.com">Richard Gauthier</a> <o:p></o:p></span></p></div><div><p class=MsoNormal><b><span style='font-size:10.0pt;font-family:"Arial",sans-serif'>To:</span></b><span style='font-size:10.0pt;font-family:"Arial",sans-serif'> <a href="mailto:general@lists.natureoflightandparticles.org" title="general@lists.natureoflightandparticles.org">Nature of Light and Particles - General Discussion</a> <o:p></o:p></span></p></div><div><p class=MsoNormal><b><span style='font-size:10.0pt;font-family:"Arial",sans-serif'>Sent:</span></b><span style='font-size:10.0pt;font-family:"Arial",sans-serif'> Thursday, July 07, 2016 6:42 AM<o:p></o:p></span></p></div><div><p class=MsoNormal><b><span style='font-size:10.0pt;font-family:"Arial",sans-serif'>Subject:</span></b><span style='font-size:10.0pt;font-family:"Arial",sans-serif'> Re: [General] double photon cycle, subjective v objective realities<o:p></o:p></span></p></div><div><p class=MsoNormal><o:p> </o:p></p></div><div><p class=MsoNormal>Chip and Grahame,<o:p></o:p></p></div><div><p class=MsoNormal> Lets be specific to the electron to avoid unnecessary vagueness. The moving electron (composed of a circulating photon) has a constant transverse internal momentum component mc and a longitudinal external momentum component p=gamma mv. These two momenta add vectorially (by the Pythagorean theorem) to give P^2 = p^2 + (mc)^2 where P=E/c is the momentum P=gamma mc of the helically circulating photon of energy E = gamma mc^2 that is the total energy of the linearly moving electron, modeled by the helically moving photon. This relationship is equivalent to the relativistic energy-momentum equation for a moving electron: E^2 = (pc)^2 + m^2 c^4 which, substituting E=Pc, gives (Pc)^2 = (pc)^2 + (mc^2) c^2 .. Dividing by c^2 gives P^2 = p^2 + (mc)^2 as given above. So as the electron speeds up, the transverse momentum component mc of the electron¡¯s total (internal plus external) momentum P remains constant even for a highly relativistic electron. The electron¡¯s constant transverse internal momentum component mc corresponds to (and leads to a derivation of) the electron¡¯s invariant mass m.<o:p></o:p></p></div><div><p class=MsoNormal> Richard<o:p></o:p></p></div><p class=MsoNormal><o:p> </o:p></p><div><blockquote style='margin-top:5.0pt;margin-bottom:5.0pt'><div><p class=MsoNormal>On Jul 6, 2016, at 10:18 AM, Dr Grahame Blackwell <<a href="mailto:grahame@starweave.com">grahame@starweave.com</a>> wrote:<o:p></o:p></p></div><p class=MsoNormal><o:p> </o:p></p><div><div><p class=MsoNormal style='background:white'><span style='font-size:10.0pt;font-family:"Arial",sans-serif;color:navy'>Yes Chip,</span><span style='font-size:9.0pt;font-family:"Helvetica",sans-serif'><o:p></o:p></span></p></div><div><p class=MsoNormal style='background:white'><span style='font-size:9.0pt;font-family:"Helvetica",sans-serif'> <o:p></o:p></span></p></div><div><p class=MsoNormal style='background:white'><span style='font-size:10.0pt;font-family:"Arial",sans-serif;color:navy'>Certainly the momentum of the confined wave increases - but that increased momentum should not ALL be reckoned as ANGULAR momentum of the electron.</span><span style='font-size:9.0pt;font-family:"Helvetica",sans-serif'><o:p></o:p></span></p></div><div><p class=MsoNormal style='background:white'><span style='font-size:9.0pt;font-family:"Helvetica",sans-serif'> <o:p></o:p></span></p></div><div><p class=MsoNormal style='background:white'><span style='font-size:10.0pt;font-family:"Arial",sans-serif;color:navy'>We know that a component of the momentum of that photon is linear - it's the linear momentum of the electron in motion. There is another component of that photon that's orthogonal to that, i.e. in the direction of the cyclic motion of the photon. As the linear velocity of the electron increases, the linear component of the photon momentum increases - however the orthogonal, cyclic, component of that photon momentum does NOT increase, since the 'pitch angle' of the helical motion of that photon increases with linear electron velocity, and so also with photon frequency, so as to precisely cancel out the effect of that increased frequency in the resolved-component cyclic direction.</span><span style='font-size:9.0pt;font-family:"Helvetica",sans-serif'><o:p></o:p></span></p></div><div><p class=MsoNormal style='background:white'><span style='font-size:9.0pt;font-family:"Helvetica",sans-serif'> <o:p></o:p></span></p></div><div><p class=MsoNormal style='background:white'><span style='font-size:10.0pt;font-family:"Arial",sans-serif;color:navy'>The angular momentum of the electron, dictated by the angular momentum contribution of the photon, does NOT depend on the FULL momentum of the photon - it ONLY depends on that component of the photon that acts cyclically, i.e. the component that's orthogonal to the linear motion of the photon. That component remains constant (as long as the radius of the photon cycle remains constant).</span><span style='font-size:9.0pt;font-family:"Helvetica",sans-serif'><o:p></o:p></span></p></div><div><p class=MsoNormal style='background:white'><span style='font-size:9.0pt;font-family:"Helvetica",sans-serif'> <o:p></o:p></span></p></div><div><p class=MsoNormal style='background:white'><span style='font-size:10.0pt;font-family:"Arial",sans-serif;color:navy'>For example, if an electron is travelling with linear speed 0.6c then its formative photon is travelling in a helical path which, if we were to flatten it out (as in relativistic energy-momentum relation) we'd find that formative photon having a linear motion component of 0.6c and cyclic speed component of 0.8c. This means that the ANGULAR momentum imparted by the photon will only be 0.8 of that which it would give if it were travelling fully cyclically at speed c (as for a static particle). Since the frequency of the photon will be increased by a gamma factor of 1/0.8 for such motion, the decreased (0.8) contribution of momentum for increased (1/0.8) frequency will be exactly what it was for the static particle.</span><span style='font-size:9.0pt;font-family:"Helvetica",sans-serif'><o:p></o:p></span></p></div><div><p class=MsoNormal style='background:white'><span style='font-size:9.0pt;font-family:"Helvetica",sans-serif'> <o:p></o:p></span></p></div><div><p class=MsoNormal style='background:white'><span style='font-size:10.0pt;font-family:"Arial",sans-serif;color:navy'>I hope that helps make things clearer.</span><span style='font-size:9.0pt;font-family:"Helvetica",sans-serif'><o:p></o:p></span></p></div><div><p class=MsoNormal style='background:white'><span style='font-size:9.0pt;font-family:"Helvetica",sans-serif'> <o:p></o:p></span></p></div><div><p class=MsoNormal style='background:white'><span style='font-size:10.0pt;font-family:"Arial",sans-serif;color:navy'>Best regards,</span><span style='font-size:9.0pt;font-family:"Helvetica",sans-serif'><o:p></o:p></span></p></div><div><p class=MsoNormal style='background:white'><span style='font-size:10.0pt;font-family:"Arial",sans-serif;color:navy'>Grahame</span><span style='font-size:9.0pt;font-family:"Helvetica",sans-serif'><o:p></o:p></span></p></div><div><p class=MsoNormal style='background:white'><span style='font-size:9.0pt;font-family:"Helvetica",sans-serif'> <o:p></o:p></span></p></div></div></blockquote></div></blockquote></div></body></html>