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<div style="direction: ltr;font-family: Tahoma;color: #000000;font-size: 10pt;">Dear Richard and everyone,
<br>
<br>
Yes, that is better. The internal momentum of the field, however, does not account for everything either. The electron rest mass is, and always was, a sum of several components. Mass from field and mass from "poincare stresses" amongst them.<br>
<br>
But there anyway remains a problem. This has been discussed, famously in the Feynmann lectures in physics in the context of field momementum and field energy - where he looks at the relationship between field momentum and field rest-mass-energy and finds a
discrepancy. This has been called the 3/4 problem or the 4/3 problem ... where e(EM) = 3/4 m c^2.<br>
<br>
You can start here .. but theref to Feynmann is better ..<br>
<br>
https://en.wikipedia.org/wiki/Electromagnetic_mass<br>
<br>
This needs fixing too.<br>
<br>
Regards, John.<br>
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<div style="direction: ltr;" id="divRpF619307"><font face="Tahoma" color="#000000" size="2"><b>From:</b> General [general-bounces+john.williamson=glasgow.ac.uk@lists.natureoflightandparticles.org] on behalf of Richard Gauthier [richgauthier@gmail.com]<br>
<b>Sent:</b> Wednesday, June 24, 2015 2:15 PM<br>
<b>To:</b> Nature of Light and Particles - General Discussion<br>
<b>Subject:</b> Re: [General] Axes plots<br>
</font><br>
</div>
<div></div>
<div>
<div class="">John W and others,</div>
<div class=""> In the spirit of rethinking fundamentals, I propose that for a single particle like an electron, “rest mass” is no longer a useful concept, rather it is even misleading, since an electron is never measured to be “at rest” . Rather for an electron,
“internal momentum” mc=0.511MeV/c is more descriptive for the invariant quantity mc in the relativistic mass-energy equation E^2 = p^2 c^2 + m^2 c^4 associated with relativistic particles. For an electron, mc = 0.511MeV/c is the magnitude of the internal momentum
of the looping or double-looping photon in our models of a “resting” electron and is invariant in all Lorentz frames, p is the “external” momentum of the electron, and E is the total energy of the electron (or the photon composing it).</div>
<div class=""> Richard</div>
<br class="">
<div>
<blockquote type="cite" class="">
<div class="">On Jun 23, 2015, at 9:33 PM, John Williamson <<a href="mailto:John.Williamson@glasgow.ac.uk" class="" target="_blank">John.Williamson@glasgow.ac.uk</a>> wrote:</div>
<br class="Apple-interchange-newline">
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<div id="divRpF367283" class="" style="direction:ltr"><font class="" face="Tahoma" size="2"><b class="">From:</b><span class="Apple-converted-space"> </span>General [<a href="mailto:general-bounces+john.williamson=glasgow.ac.uk@lists.natureoflightandparticles.org" class="" target="_blank">general-bounces+john.williamson=glasgow.ac.uk@lists.natureoflightandparticles.org</a>]
on behalf of Andrew Meulenberg [<a href="mailto:mules333@gmail.com" class="" target="_blank">mules333@gmail.com</a>]<br class="">
<b class="">Sent:</b><span class="Apple-converted-space"> </span>Tuesday, June 23, 2015 12:17 PM<br class="">
<b class="">To:</b><span class="Apple-converted-space"> </span>Nature of Light and Particles - General Discussion<br class="">
<b class="">Cc:</b><span class="Apple-converted-space"> </span>Nick Bailey; Anthony Booth; Manohar .; Ariane Mandray; David Mathes<br class="">
<b class="">Subject:</b><span class="Apple-converted-space"> </span>Re: [General] Axes plots<br class="">
</font><br class="">
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<div class=""></div>
<div class="">
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<div class="">Dear John,<br class="">
<br class="">
</div>
We need to get this group together for a week or more of discussions. I have so much to learn. Each answer produces more questions. I am afraid not to ask them because they may not come up again and I will have lost an opportunity for obtaining important information.<br class="">
<ol class="">
<li class="">the 'quarter turn' would probably require a whole lecture to bring me up to speed. It may relate to 'up' and 'down' in QM being orthogonal rather than along the same axis.<span class="Apple-converted-space"> </span><font class="" color="0000FF">Explained
in other email</font><br class="">
</li><li class="">I agree with "integer wavelengths to minimise energy." However, is this true for the case of the 'untwisted' circularly-polarized photon?<span class="Apple-converted-space"> </span><font class="" color="0000FF">It- too is twisted - it is not turned
though</font><br class="">
</li><li class="">Aligning superposed spin axes of in-phase photons should increase the field energies and therefore the total energy?<span class="Apple-converted-space"> </span><font class="" color="0000FF">I mean aligning the axes of twist and turn - not aligning
two photons.</font><br class="">
</li></ol>
<div class="" style="margin-top:0px; margin-bottom:0px">This is much more interesting than writing papers or creating exam questions. However, we do have those obligations.<span class="Apple-converted-space"> </span><font class="" color="0000FF">Don't remind
me - I'm on holiday remember!</font><br class="">
</div>
<p class="" style="margin-top:0px; margin-bottom:0px"></p>
<div class="" style="margin-top:0px; margin-bottom:0px">Andrew</div>
<div class="" style="margin-top:0px; margin-bottom:0px">__________________<br class="">
</div>
<div class="">
<div class="">
<div class="gmail_extra"><br class="">
<div class="gmail_quote">On Tue, Jun 23, 2015 at 6:35 AM, John Williamson<span class="Apple-converted-space"> </span><span dir="ltr" class=""><<a href="mailto:John.Williamson@glasgow.ac.uk" class="" target="_blank">John.Williamson@glasgow.ac.uk</a>></span><span class="Apple-converted-space"> </span>wrote:<br class="">
<blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex; border-left-width:1px; border-left-style:solid; border-left-color:rgb(204,204,204); padding-left:1ex">
<div class="">
<div class="" style="direction:ltr; font-family:Tahoma; font-size:10pt">Hi Andrew,<br class="">
<br class="">
No. The twist comes form the base equations - its not a "knob" but a consequence of the underlying thing the differential equations are describing. Each differential is a (precise) quarter turn in some space. Further, you need to join at integer wavelengths
to minimise energy. Higher twist objects (2 - 3 ans so on) also need to minimise energy. Physically this means aligning spin axes. there are actually only so many real dimensions to go round (despite the idiocies of string theory).<br class="">
<br class="">
Regards, John.<br class="">
<div class="" style="font-family:'Times New Roman'; font-size:16px">
<hr class="">
<div class="" style="direction:ltr"><font class="" face="Tahoma" size="2"><b class="">From:</b><span class="Apple-converted-space"> </span>General [general-bounces+john.williamson=<a href="mailto:glasgow.ac.uk@lists.natureoflightandparticles.org" class="" target="_blank">glasgow.ac.uk@lists.natureoflightandparticles.org</a>]
on behalf of Andrew Meulenberg [<a href="mailto:mules333@gmail.com" class="" target="_blank">mules333@gmail.com</a>]<br class="">
<b class="">Sent:</b><span class="Apple-converted-space"> </span>Tuesday, June 23, 2015 11:28 AM<br class="">
<b class="">To:</b><span class="Apple-converted-space"> </span>Nature of Light and Particles - General Discussion<br class="">
<b class="">Cc:</b><span class="Apple-converted-space"> </span>Nick Bailey; Anthony Booth; Manohar .; Ariane Mandray; David Mathes<br class="">
<b class="">Subject:</b><span class="Apple-converted-space"> </span>Re: [General] Axes plots<br class="">
</font><br class="">
</div>
<div class="">
<div class="h5">
<div class=""></div>
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<div dir="ltr" class="">
<div class="">
<div class="">Dear John W. and all,<br class="">
<br class="">
</div>
If you start with a circularly-polarized photon, give it one full twist per wavelength (to 'unwind' it), and then close it at any point, could you not create the out-going<span class="Apple-converted-space"> </span><b class="">E</b>-field that we attribute
to the electron or positron? However, since there is no reason to join the twisted photon only at integer wavelengths, then the electron would not have fixed properties. The only thing fixed would be the amount of twist per wavelength. What causes the photon
twist in any of the models and, if the above model is correct, why don't we have particles with a continuum of charge (and spin?) values?<br class="">
<br class="">
</div>
Andrew<br class="">
<div class="">
<div class="">__________________________________<br class="">
<br class="">
<div class="gmail_extra"><br class="">
<div class="gmail_quote">On Tue, Jun 23, 2015 at 12:19 AM, John Williamson<span class="Apple-converted-space"> </span><span dir="ltr" class=""><<a href="mailto:John.Williamson@glasgow.ac.uk" class="" target="_blank">John.Williamson@glasgow.ac.uk</a>></span><span class="Apple-converted-space"> </span>wrote:<br class="">
<blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex; border-left-width:1px; border-left-style:solid; border-left-color:rgb(204,204,204); padding-left:1ex">
<div class="">
<div class="" style="direction:ltr; font-family:Tahoma; font-size:10pt">Here are some using a set of axes, as in the POS paper circulated earlier<br class="">
<br class="">
Torfieldrock just shows three views of a positron.<span class="Apple-converted-space"> </span><br class="">
<br class="">
Torfieldsinglemin0.4 is just a simple double loop. Not too busy.<br class="">
<br class="">
Trockrevert is similar to the first, but with a glass torus.<br class="">
<br class="">
Cheers, John.<br class="">
</div>
</div>
<br class="">
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