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<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US">Ok Richard, I will try,</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US"><br>
</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US">Gentlemen, I think you are entirely missing the point about (quantum)
spin. It is not primarily about a value. It is primarily about the (experimental) fact that it comes in two varieties “up” or “down”. These two states follow an exclusion principle.
<br>
</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US"><br>
</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US">In other words the spin of an elementary electron or proton is in no
way like a little spinning top. Forget about it. Not even close. In no way an analogy.
<br>
</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US"><br>
</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US">It is BOTH far far smaller (pretty much zero) and far far bigger than
that, of which more later.</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US">Hopefully, this does not come as a complete surprise. You all already
know this to be true.</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US">Now the important experimental stuff for spin comes from (at least)
four different fields. I cannot teach you all (or any!) these in an email. In fact, aspects of these I could not teach you at all as I do not master all of them and some of you know more about one or two of them than do I. Ignorance knows (little or) no bounds!
However I am (supposed to be) expert in a couple of the relevant fields so - with that disclaimer here goes …</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US"><br>
</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US">Firstly, the strongest evidence (as far as I know – this is not my field)
for the value and the coupling of the photon spin and electron spin comes from atomic and molecular physics and the “allowed” transitions emitting and absorbing photons.<span style="mso-spacerun:yes">
</span>The limiting “spin” of the electron and photon is strongly related, in that (if one accepts angular momentum conservation) one is half (or double) the other. The analogy is a spin up” system (+1/2) emits a spin 1 photon (+1) flipping to spin down (-1/2).
The photon is subsequently absorbed by a spin “down” which flips to a spin “up”. This is just a sketch, and of course lots of transitions which are not “allowed” do happen – but they have probabilities which are much lower. Anyway, in this context it
<span style="mso-spacerun:yes"> </span>could look as though “spin” was a purely electromagnetic phenomenon. This would be the world-view one came away with if this was all one knew.</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US"><br>
</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US">The big surprise is, if one steps out of this field and looks at others,
that spin is not just characteristic of the electromagnetic “strength”. It is strong. It is stronger than strong!</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US">Before discussing the evidence for that note that the experimental properties
of spin have been the starting point for many beautiful theories. Alex, for example, has used this. In this he has been following Kerr, Newman Carter and others – all beautiful and short! papers. He argued in 1973 that if the spin density is what it is there
must be some strong force at play. He and others have ascribed this to gravitation, within the framework of GR. Though others, Carter for example, attribute this to a breakdown of the theory of gravitation. Whatever the “force” confining the spin is it certainly
large. This is not to mention the (infinitely) large forces one needs to contain a point charge. This is a point where Alex and I disagree (though it is only in a name) – for me gravitation is a weak thing – I would just prefer to call the strong stuff something
else to avoid confusion.<span style="mso-spacerun:yes"> </span>Effectively, Martin and also started from this in our 1997 paper, noting that if the initial photon was spinning at one unit, the resultant double-loop electron spun with half a unit.</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US"><br>
</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US">Anyway, this is the task at hand, gentlemen, one needs (eventually)
to explain all of this simultaneously, not just bits of it. <br>
</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US"><br>
</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US">Another beautiful, and more recent theory in another field also takes
spin as a starting point. This is Carver Mead’s “Collective electrodynamics”. Also very beautiful and I recommend you all read it. He takes spin as a central paradigm in solid state physics and derives much of value and beauty from this.</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US"><br>
</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US">However I diverge a bit. Lets get back to what is known from experiment,
and in particular with what happens at high energy.</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US"><br>
</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US">This is completely astonishing.</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US"><br>
</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US">Firstly one must realize that there is no difference in the energy of
an elementary particle for spin “up” or “down”. I do not mean “no difference” figuratively here. The energy difference is zero. One only observes a difference at all in the presence of an external magnetic field – and this is a tiny tiny fraction of the mass-energy
of the particles concerned. Now one can prepare a set of particles in a given spin state rather easily electromagnetically – and with very low energy input (like zero!). If one does this, and subsequently accelerates the spin up or spin down particles while
maintaining that spin and then bash them into one another, things at least as amazing as what happened in Rutherfords experiment happens. One expects very little at the scale of the strong interaction. One has only inputted a little eentsy bit of energy into
the spin component far away and long ago. All gone now. For this reason one does not really measure the (tiny) spin of an elementary particle as a number in HEP – one observes its effects – and these are astonishingly large.</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US"><br>
</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US">What actually happens in collisions is MONSTROUS.</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US"><br>
</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US">This was first done at the zero gradient synchrotron at Argonne (the
clue of maintaining the spin is in the name). Look at Scientific American May 1979. Nice pictures. Shocking. Disproves the quark model! Oh dear.</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US"><br>
</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US">Take the probability of scattering of unpolarised protons on protons
as a base. Call it 1. Remember this scattering is dominated by the strong interaction at high energies. At 4-momentum transfers of the order of the proton mass the spin parallel probabilty is about 2 and for spin antiparallel about a half.<span style="mso-spacerun:yes">
</span>It is as if one had wrapped ones tank in blue cling film – and discovered that shells then passed right through it half the time, whereas if one used red cling film the shells tended to blow it up, even if they missed it a bit. Further, if one tries
to decompose the results in terms of the internal parton (quark) spin it does not work at all. The quark model breaks down. It was largely the complete refusal of the HEP community to accept this experimental fact and chase it down that led me to leave it
in disgust in the mid eighties.</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US"><br>
</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US">Experimentally, more astonishing still one finds a beautiful resonance,
a little below the proton mass-energy, where the protons become even more transparent. Utterly astonishing (and still completely unexplained) behavior. Look at it! Quantum spin is crucial. The “exclusion principle” is not infinitely potent, but remains potent
up to energies exceeding the total mass-energy of the elementary particles themselves. Further, there is very clear structure (in energy) in its potency. Explain that! We do in terms of the double-loop model – see below.</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US"><br>
</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US">Subsequent experiments using polarised targets and polarized beams all
found such kinds of behavior. Conclusion: whatever spin is it is involved with generating interactions as strong or stronger than the so-called strong interaction. Conventionally, one ascribes these to “exchange” forces though the detail of what is supposedly
exchanged in the framework (of QCD, for example) throws up serious contradictions in the detail. Normally, this would have meant that one threw away ones theory and came up with something better, but the scientific method seems to have gone out of fashion
in these circles and this has yet to happen (to my knowledge).</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US"><br>
</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US">Coming back to our discussion. Even if one could line up the whole electron
or proton as a pure spinning top, oriented with spin up (in the classical sense) one would not observe anything like this. The energy would be limited to the energy in the spin. One sees much more. These particles do not behave AT ALL like little spinning
tops. NO way. Not a bit. Forget about it!</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US"><br>
</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US">In the context of solid state physics one also sees astonishing effects
associated with spin. Mad magnets. Spin blockade in quantum dots. Ridiculously high Tc superconductors. Giant magneto-resistance. Lots more “exchange” stuff way beyond standard QM or EM.</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US"><br>
</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US">Conclusion: a proper understanding of the underlying nature of quantum
spin will prove crucial in developing a proper theory of how it all works. <br>
</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US"><br>
</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US">Martin and I have explained these effects with our models in terms of
the interference of the internal fields of the elementary particles – doubling the mass-energy if the interference is precisely in phase (leading to an exclusion) and giving zero energy if in antiphase. We talked about this in our early work leading to the
1997 paper. I talked about this at CYBCOM in 2008 and presented a paper arguing this in 2012 (proceedings of MENDEL 2012). This is how one makes fermions from bosons – tie them up into a double loop. Martin has been looking at this in more detail in the context
of much weaker overlaps in the solid state. <br>
</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US"><br>
</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US">So – if one looks one sees effects larger even than the E = hbar nu
one would ascribe to the full mass energy of the elementary particles. This is something I should have mentioned before. In the formula above hbar takes the role of an elementary, quantized angular momentum. That is the angular momentum (spin energy) accounts
for ALL the energy of the photon. In this sense, for a circularly polarized photon, all of its energy is rotational. This does not mean that the photon is rotating like a little massive ball on a stick. Not at all. No ball. No stick.<span style="mso-spacerun:yes">
</span>It is the EM field that carries the rotation. The rotation appears at its simplest, as the differential of the (perpendicular) differential of a (perpendicular) vector.<span style="mso-spacerun:yes">
</span>More properly, one may wish to see it (as measured in experiment) as the differential of one vector times the differential of the differential of another (at least another phase (loop)). I said before the rotation “axis” must be viewed as at least complex,
and in my view more complex than complex (complex is only 2D and one needs at least 4 dimensions to describe it dynamically)
<span style="mso-spacerun:yes"> </span>to give any hope of explaining the body of experimental facts. Three complex vectors would do it (for example). Cheers Alex – but I think one could make it simpler that that!
<br>
</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US"><br>
</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US">Gentlemen, coming back to the discussion about the spin of the moving
electron -if we take angular momentum to be conserved then the spin of an electron passing you fast is exactly the same as one passing you slow as JD said as – think about it – you have not spun it up or down by accelerating yourself. One may wish to try to
explain this in terms of the spin of tops, relativistic or otherwise (also conserved) but one is then not getting into a discussion about the really interesting stuff about spin at all.</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US"><br>
</span></p>
<p class="MsoNormal" style="mso-margin-top-alt:auto;mso-margin-bottom-alt:auto"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman"" lang="EN-US">Regards, John.</span></p>
<div style="font-family: Times New Roman; color: #000000; font-size: 16px">
<hr tabindex="-1">
<div style="direction: ltr;" id="divRpF890171"><font face="Tahoma" color="#000000" size="2"><b>From:</b> General [general-bounces+john.williamson=glasgow.ac.uk@lists.natureoflightandparticles.org] on behalf of John Duffield [johnduffield@btconnect.com]<br>
<b>Sent:</b> Tuesday, July 12, 2016 7:23 PM<br>
<b>To:</b> 'Nature of Light and Particles - General Discussion'<br>
<b>Subject:</b> Re: [General] double photon cycle, subjective v objective realities<br>
</font><br>
</div>
<div></div>
<div>
<div class="WordSection1">
<p class="MsoNormal"><span style="font-size:11.0pt; font-family:"Calibri",sans-serif; color:#1F497D"> </span></p>
<p class="MsoNormal"><span style="font-size:14.0pt; font-family:"Calibri",sans-serif; color:#1F497D">Can I chip in to say that IMHO Compton scattering takes a “slice” off the photon and gives it to the electron in an asymmetrical fashion. As a result, the
electron moves. It moves because it’s a circulating photon that’s no longer a <i>
symmetrical</i> circulating photon. It’s hard to visualize this, but simplify the electron to a photon going round in a circular path. When Compton scattering occurs, energy is added so the wavelength reduces, but asymmetrically. It’s like drawing say 355 degrees
of a circle, then without lifting your pen, drawing another 355 degrees of a circle, and so on:</span></p>
<p class="MsoNormal"><span style="font-size:14.0pt; font-family:"Calibri",sans-serif; color:#1F497D"> </span></p>
<p class="MsoNormal"><span style="font-size:14.0pt; font-family:"Calibri",sans-serif; color:#1F497D"><img id="Picture_x0020_1" src="cid:image001.jpg@01D1DC72.3EBF4540" height="237" width="414"></span><span style="font-size:14.0pt; font-family:"Calibri",sans-serif; color:#1F497D">
</span></p>
<p class="MsoNormal"><span style="font-size:14.0pt; font-family:"Calibri",sans-serif; color:#1F497D"> </span></p>
<p class="MsoNormal"><span style="font-size:14.0pt; font-family:"Calibri",sans-serif; color:#1F497D">As for the exact details of what happens with a fast-moving electron, I’m not sure. I am reminded of extending a slinky, but I know that an electron doesn’t
change just because I move past it fast. And I wish that all physicists only had that to disagree upon.
</span></p>
<p class="MsoNormal"><span style="font-size:14.0pt; font-family:"Calibri",sans-serif; color:#1F497D"> </span></p>
<p class="MsoNormal"><span style="font-size:14.0pt; font-family:"Calibri",sans-serif; color:#1F497D">Regards</span></p>
<p class="MsoNormal"><span style="font-size:14.0pt; font-family:"Calibri",sans-serif; color:#1F497D">JohnD</span></p>
<p class="MsoNormal"><span style="font-size:11.0pt; font-family:"Calibri",sans-serif; color:#1F497D"> </span></p>
<div>
<div style="border:none; border-top:solid #E1E1E1 1.0pt; padding:3.0pt 0cm 0cm 0cm">
<p class="MsoNormal"><b><span style="font-size:11.0pt; font-family:"Calibri",sans-serif" lang="EN-US">From:</span></b><span style="font-size:11.0pt; font-family:"Calibri",sans-serif" lang="EN-US"> General [mailto:general-bounces+johnduffield=btconnect.com@lists.natureoflightandparticles.org]
<b>On Behalf Of </b>Chip Akins<br>
<b>Sent:</b> 12 July 2016 16:25<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</span></p>
</div>
</div>
<p class="MsoNormal"> </p>
<p class="MsoNormal"><span style="color:black" lang="EN-US">Hi Again Grahame</span></p>
<p class="MsoNormal"><span style="color:black" lang="EN-US"> </span></p>
<p class="MsoNormal"><span style="color:black" lang="EN-US">One issue I have been contemplating is the “spin mode” of the energy in the photon and the electron.</span></p>
<p class="MsoNormal"><span style="color:black" lang="EN-US"> </span></p>
<p class="MsoNormal"><span style="color:black" lang="EN-US">You said…”</span><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy">[A ‘quick fix’ would of course be to propose a linearly polarized photon with zero spin – i.e. 50/50 superposition
of left and right circularly polarized elements. This, though, is rather a cop-out as it removes a possible explanation for other electron features, notably charge; it also doesn’t feel right.]</span><span style="color:black" lang="EN-US">”</span></p>
<p class="MsoNormal"><span style="color:black" lang="EN-US"> </span></p>
<p class="MsoNormal"><span style="color:black" lang="EN-US">But perhaps that is not a “quick fix” at all. It may well be that characterizing the electron as a confined photon is actually a quick fix and that the energy in the electron is configured in a completely
different “spin mode” than that of the photon. We already know that the electron spin is, at the very least, different than the photon spin, so we might be trying to force fit the photon solution where it does not apply at all (except that the energy can be
configured in two completely different stable modes).</span></p>
<p class="MsoNormal"><span style="color:black" lang="EN-US"> </span></p>
<p class="MsoNormal"><span style="color:black" lang="EN-US">Thoughts?</span></p>
<p class="MsoNormal"><span style="color:black" lang="EN-US"> </span></p>
<p class="MsoNormal"><span style="color:black" lang="EN-US">Chip</span></p>
<p class="MsoNormal"><span style="color:black" lang="EN-US"> </span></p>
<div>
<div style="border:none; border-top:solid #E1E1E1 1.0pt; padding:3.0pt 0cm 0cm 0cm">
<p class="MsoNormal"><b><span style="font-size:11.0pt; font-family:"Calibri",sans-serif" lang="EN-US">From:</span></b><span style="font-size:11.0pt; font-family:"Calibri",sans-serif" lang="EN-US"> General [<a href="mailto:general-bounces+chipakins=gmail.com@lists.natureoflightandparticles.org" target="_blank">mailto:general-bounces+chipakins=gmail.com@lists.natureoflightandparticles.org</a>]
<b>On Behalf Of </b>Dr Grahame Blackwell<br>
<b>Sent:</b> Tuesday, July 12, 2016 8:24 AM<br>
<b>To:</b> Nature of Light and Particles - General Discussion <<a href="mailto:general@lists.natureoflightandparticles.org" target="_blank">general@lists.natureoflightandparticles.org</a>><br>
<b>Subject:</b> Re: [General] double photon cycle, subjective v objective realities</span></p>
</div>
</div>
<p class="MsoNormal"><span lang="EN-US"> </span></p>
<div>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy">Hi Chip, Richard, John W,</span></p>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy"> </span></p>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy">I think we all realised, from Chip’s email if not before, that there’s an issue to be addressed with regard to a photon’s angular momentum in the overall behaviour
of a photon-formed electron.</span></p>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy"> </span></p>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy">It seems to me that, whatever photon-based electron model any one of us chooses to put forward, when that electron is in motion then there will be a component of
that formative photon’s angular momentum in the direction of motion of the electron (i.e. about any axis in that direction). More than this, my preliminary investigations suggest that it would take a very creative model indeed to ensure that the rise in that
component with increase in electron speed would be exactly balanced by a drop-off in the component from the photon’s linear momentum to give the electron a constant angular momentum.</span></p>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy"> </span></p>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy">[A ‘quick fix’ would of course be to propose a linearly polarized photon with zero spin – i.e. 50/50 superposition of left and right circularly polarized elements.
This, though, is rather a cop-out as it removes a possible explanation for other electron features, notably charge; it also doesn’t feel right.]</span></p>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy"> </span></p>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy">I believe we’re all agreed that:</span></p>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy">(a) There’s more than enough evidence to confirm that the concept of an electron (and likely also other elementary particles) being formed from a closed-loop photon
is totally valid;</span></p>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy">(b) Formation of an electron involves a double-loop per wavelength of the photon, at least for the static electron. All else aside this is clearly indicated by
zitterbewegung.</span></p>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy"> </span></p>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy">I’m guessing we all also agree on the validity of the so-called Relativistic Energy-Momentum Relation (whether or not we subscribe to the idea of objective frame
symmetry). There are quite a few points that can be drawn from that, as I see it.</span></p>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy"> </span></p>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy">Most importantly, the REMR represents the full momentum of a moving electron (i.e. momentum of its formative photon) in terms of linear and cyclic components. Expressed
diagrammatically, these three components form a right-angled triangle that defines the relative directions of the instantaneous velocity components (linear, cyclic, overall) of that photon (since these must necessarily follow momentum component directions).</span></p>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy"> </span></p>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy">It’s pretty clear, first of all, that if the linear velocity component is v (as it is) and the overall velocity is anything other than c (in directions as given
by REMR) then the cyclic velocity component will not be orthogonal to the linear component (as it must be). From this I believe that we can confirm that the velocity (at least the mean velocity) of an electron-forming photon must be c; I think this rules
out certain proposals.</span></p>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy"> </span></p>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy">Secondly, that velocity triangle gives cyclic velocity component as c/gamma. Those who subscribe to SR’s objective frame symmetry would presumably expect the double-loop
to complete in a time gamma tau, where tau is the time for that double-loop for a static electron (since from the static perspective that double-loop in the moving electron would have to correspond with the time-dilated interval in the moving frame). I also
see the double-loop completing in that time, since I regard energy flow as the underlying mechanism driving the passage of time; a slower rate of time-experience (time dilation) is the consequence, rather than the cause, of that reduced looping rate.</span></p>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy"> </span></p>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy">Whichever is the case, a looping rate reduced by a factor gamma and achieved by a flow speed component also reduced by a factor gamma indicates a constant path-length,
i.e. a constant radius for the cyclic path of the formative photon. This appears to be an inescapable conclusion from consideration of the REMR and time dilation.</span></p>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy"> </span></p>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy">[For completeness I should add that SR frame symmetry requires that each double-loop is also seen as exactly one full single wavelength from within the moving frame;
for me this raises an irreconcilable contradiction in SR.]</span></p>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy"> </span></p>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy">Back to that photon spin</span></p>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy">==================</span></p>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy"> </span></p>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy">Clearly either electron spin increases with speed of the electron’s linear motion – or it doesn’t. If it doesn’t then this implies some aspect of quantum mechanics
that needs further consideration. Rather more significantly for us, I believe it also rules out the whole concept of electrons being formed from looping photons.</span></p>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy"> </span></p>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy">That last seemingly outrageous statement follows from the tendency towards flat-lining of the formative photon as an electron tends towards speed c. Unless we consider
that photon to be other than circularly polarized – which I believe raises serious difficulties with other aspects of the model – then this means that the electron’s angular momentum in its direction of motion tends to at least hbar – which is clearly inconsistent
with constancy of electron angular momentum with increasing speed.</span></p>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy"> </span></p>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy">Rather less problematic (as I see it) is the notion that the electron’s angular momentum in its direction of motion increases with its speed. I don’t know of any
experimental evidence showing conclusively that this is not the case, if others do then of course that would be of interest.</span></p>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy"> </span></p>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy">Obviously if angular momentum does increase then that must be by virtue of its being transferred from elsewhere. Since increase in velocity must be caused by an
input of energy – a real or virtual photon – then the most obvious course is to consider a Compton scattering event that increases the velocity of the electron.</span></p>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy"> </span></p>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy">It’s known that in general Compton scattering leads to a change in polarisation state of the scattered photon. From what little I’ve seen, such changes have been
calculated from theory and confirmed in principle by experiment; that theory doesn’t generally include the notion of an electron’s angular momentum varying with speed, as far as I know. That (likely) omission would make negligible difference in all but the
most extreme case: increase of electron speed from sub-relativistic to highly relativistic in a single step – since the change in scattered-photon spin would correspond to the change in v/c for the electron.</span></p>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy"> </span></p>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy">To summarise: The concept of an electron formed from a circularly-polarised photon looping at constant radius for all speeds of the electron appears to be consistent
with all experimental evidence, other than maybe definitive evidence on electron radius at speed* (unless electrons moving at high speed have been shown, and not just inferred, to have spin ½); this observation is based on the assumption that no evidence exists
of photon polarisation state changes in high-energy Compton scattering events with sufficient powers of discrimination (i.e. accuracy) to definitively show that no angular momentum has been passed from the photon to the electron, other than that accounted
for by a change in direction of motion of the electron.</span></p>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy">* I have yet to look at this.</span></p>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy"> </span></p>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy">Of course this is just my view, based on my understanding of available scientific data. I’d be interested to hear other views on these observations</span></p>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy"> </span></p>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy">[Richard, I hope it’s clear from the above why I have reservations over your proposed v=sqrt(2)c spin-1/2 photon model of the electron. In particular I can’t see
how that model can be reconciled with the Relativistic Energy-Momentum Relation in terms of correspondence of directions for components of momentum and velocity.]</span></p>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy"> </span></p>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy">Regards to all,</span></p>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy">Grahame</span></p>
</div>
<blockquote style="border:none; border-left:solid navy 1.5pt; padding:0cm 0cm 0cm 4.0pt; margin-left:3.75pt; margin-top:5.0pt; margin-right:0cm; margin-bottom:5.0pt">
<div>
<p class="MsoNormal"><span style="font-size:10.0pt; font-family:"Arial",sans-serif" lang="EN-US">----- Original Message -----
</span></p>
</div>
<div>
<p class="MsoNormal" style="background:#E4E4E4"><b><span style="font-size:10.0pt; font-family:"Arial",sans-serif" lang="EN-US">From:</span></b><span style="font-size:10.0pt; font-family:"Arial",sans-serif" lang="EN-US">
<a href="mailto:richgauthier@gmail.com" title="richgauthier@gmail.com" target="_blank">
Richard Gauthier</a> </span></p>
</div>
<div>
<p class="MsoNormal"><b><span style="font-size:10.0pt; font-family:"Arial",sans-serif" lang="EN-US">To:</span></b><span style="font-size:10.0pt; font-family:"Arial",sans-serif" lang="EN-US">
<a href="mailto:general@lists.natureoflightandparticles.org" title="general@lists.natureoflightandparticles.org" target="_blank">
Nature of Light and Particles - General Discussion</a> </span></p>
</div>
<div>
<p class="MsoNormal"><b><span style="font-size:10.0pt; font-family:"Arial",sans-serif" lang="EN-US">Sent:</span></b><span style="font-size:10.0pt; font-family:"Arial",sans-serif" lang="EN-US"> Friday, July 08, 2016 6:13 AM</span></p>
</div>
<div>
<p class="MsoNormal"><b><span style="font-size:10.0pt; font-family:"Arial",sans-serif" lang="EN-US">Subject:</span></b><span style="font-size:10.0pt; font-family:"Arial",sans-serif" lang="EN-US"> Re: [General] double photon cycle, subjective v objective realities</span></p>
</div>
<div>
<p class="MsoNormal"><span lang="EN-US"> </span></p>
</div>
<div>
<p class="MsoNormal"><span lang="EN-US">Hello Grahame,</span></p>
</div>
<div>
<p class="MsoNormal"><span lang="EN-US"> </span></p>
</div>
<div>
<p class="MsoNormal"><span lang="EN-US"> Unfortunately the situation is not so simple. Neither of our published electron models includes a specific photon model with its own spin, where this photon model moves along the helical trajectory described in our
models. If that photon moving along the helical trajectory has a spin that is is independent of the energy of the photon (which is the nature of photons) then as the photon's trajectory in the your double-loop constant helical radius electron model gets more
and more straight with increasing electron speed, then the spin of this circulating photon adds more and more to the spin 1/2 of your electron model produced by its circling transverse component of momentum mc at constant radius R. The result is that a circulating
spin 1 photon along your constant radius R helical trajectory would give your electron model a total spin of one and a half units of spin hbar at highly relativistic velocities. A circulating spin 1/2 photon traveling along your constant radius R trajectory
would give your electron model a total spin of 1/2 + 1/2 = 1 unit of hbar of spin at highly relativistic velocities. It is only if the radius R of the photon’s helical trajectory decreases with increasing velocity to become insignificant (compared to R in
a resting electron) at relativistic velocities that the spin of the electron model at relativistic velocities will equal only the spin of the photon composing the electron model. Ideally the helically circulating photon model of the electron will have longitudinal
spin component 1/2 hbar at all velocities of the electron model from very slow velocities to very highly relativistic velocities. </span></p>
</div>
<div>
<p class="MsoNormal"><span lang="EN-US"> </span></p>
</div>
<div>
<p class="MsoNormal"><span lang="EN-US"> I have an unpublished internally superluminal (v=c sqrt(2) ) helically circulating spin-1/2 photon model whose spin remains 1/2 at all energies, which may be suitable for modeling the electron. I described this photon
model in this email list in the past. The radius of my published spin-1/2 charged-photon electron model’s photon trajectory decreases as 1/gamma^2 with increasing electron velocity, so this does not produce the complication described above when the helical
radius of the photon’s trajectory is a constant R at all electron velocities.</span></p>
</div>
<div>
<p class="MsoNormal"><span lang="EN-US"> </span></p>
</div>
<div>
<p class="MsoNormal"><span lang="EN-US"> Richard</span></p>
</div>
<p class="MsoNormal"><span lang="EN-US"> </span></p>
<div>
<blockquote style="margin-top:5.0pt; margin-bottom:5.0pt">
<div>
<p class="MsoNormal"><span lang="EN-US">On Jul 7, 2016, at 1:00 AM, Dr Grahame Blackwell <<a href="mailto:grahame@starweave.com" target="_blank">grahame@starweave.com</a>> wrote:</span></p>
</div>
<p class="MsoNormal"><span lang="EN-US"> </span></p>
<div>
<div>
<p class="MsoNormal" style="background:white"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy" lang="EN-US">Thanks Richard,</span><span style="font-size:9.0pt; font-family:"Helvetica",sans-serif" lang="EN-US"></span></p>
</div>
<div>
<p class="MsoNormal" style="background:white"><span style="font-size:9.0pt; font-family:"Helvetica",sans-serif" lang="EN-US"> </span></p>
</div>
<div>
<p class="MsoNormal" style="background:white"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy" lang="EN-US">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><span style="font-size:9.0pt; font-family:"Helvetica",sans-serif" lang="EN-US"></span></p>
</div>
<div>
<p class="MsoNormal" style="background:white"><span style="font-size:9.0pt; font-family:"Helvetica",sans-serif" lang="EN-US"> </span></p>
</div>
<div>
<p class="MsoNormal" style="background:white"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy" lang="EN-US">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><span style="font-size:9.0pt; font-family:"Helvetica",sans-serif" lang="EN-US"></span></p>
</div>
<div>
<p class="MsoNormal" style="background:white"><span style="font-size:9.0pt; font-family:"Helvetica",sans-serif" lang="EN-US"> </span></p>
</div>
<div>
<p class="MsoNormal" style="background:white"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy" lang="EN-US">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<span class="apple-converted-space"> </span><em><span style="font-family:"Arial",sans-serif">not</span></em><span class="apple-converted-space"> </span>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><span style="font-size:9.0pt; font-family:"Helvetica",sans-serif" lang="EN-US"></span></p>
</div>
<div>
<p class="MsoNormal" style="background:white"><span style="font-size:9.0pt; font-family:"Helvetica",sans-serif" lang="EN-US"> </span></p>
</div>
<div>
<p class="MsoNormal" style="background:white"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy" lang="EN-US">Best regards,</span><span style="font-size:9.0pt; font-family:"Helvetica",sans-serif" lang="EN-US"></span></p>
</div>
<div>
<p class="MsoNormal" style="background:white"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy" lang="EN-US">Grahame</span><span style="font-size:9.0pt; font-family:"Helvetica",sans-serif" lang="EN-US"></span></p>
</div>
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<p class="MsoNormal" style="background:white"><span style="font-size:10.0pt; font-family:"Arial",sans-serif" lang="EN-US">----- Original Message -----<span class="apple-converted-space"> </span></span></p>
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<p class="MsoNormal" style="background:#E4E4E4"><b><span style="font-size:10.0pt; font-family:"Arial",sans-serif" lang="EN-US">From:</span></b><span class="apple-converted-space"><span style="font-size:10.0pt; font-family:"Arial",sans-serif" lang="EN-US"> </span></span><span style="font-size:10.0pt; font-family:"Arial",sans-serif" lang="EN-US"><a href="mailto:richgauthier@gmail.com" title="richgauthier@gmail.com" target="_blank">Richard
Gauthier</a></span></p>
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<p class="MsoNormal" style="background:white"><b><span style="font-size:10.0pt; font-family:"Arial",sans-serif" lang="EN-US">To:</span></b><span class="apple-converted-space"><span style="font-size:10.0pt; font-family:"Arial",sans-serif" lang="EN-US"> </span></span><span style="font-size:10.0pt; font-family:"Arial",sans-serif" lang="EN-US"><a href="mailto:general@lists.natureoflightandparticles.org" title="general@lists.natureoflightandparticles.org" target="_blank">Nature
of Light and Particles - General Discussion</a></span></p>
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<p class="MsoNormal" style="background:white"><b><span style="font-size:10.0pt; font-family:"Arial",sans-serif" lang="EN-US">Sent:</span></b><span class="apple-converted-space"><span style="font-size:10.0pt; font-family:"Arial",sans-serif" lang="EN-US"> </span></span><span style="font-size:10.0pt; font-family:"Arial",sans-serif" lang="EN-US">Thursday,
July 07, 2016 6:42 AM</span></p>
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<p class="MsoNormal" style="background:white"><b><span style="font-size:10.0pt; font-family:"Arial",sans-serif" lang="EN-US">Subject:</span></b><span class="apple-converted-space"><span style="font-size:10.0pt; font-family:"Arial",sans-serif" lang="EN-US"> </span></span><span style="font-size:10.0pt; font-family:"Arial",sans-serif" lang="EN-US">Re:
[General] double photon cycle, subjective v objective realities</span></p>
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<p class="MsoNormal" style="background:white"><span style="font-size:9.0pt; font-family:"Helvetica",sans-serif" lang="EN-US">Chip and Grahame,</span></p>
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<p class="MsoNormal" style="background:white"><span style="font-size:9.0pt; font-family:"Helvetica",sans-serif" lang="EN-US"> 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.</span></p>
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<p class="MsoNormal" style="background:white"><span style="font-size:9.0pt; font-family:"Helvetica",sans-serif" lang="EN-US"> Richard</span></p>
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<p class="MsoNormal" style="background:white"><span style="font-size:9.0pt; font-family:"Helvetica",sans-serif" lang="EN-US">On Jul 6, 2016, at 10:18 AM, Dr Grahame Blackwell <<a href="mailto:grahame@starweave.com" target="_blank">grahame@starweave.com</a>>
wrote:</span></p>
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<p class="MsoNormal" style="background:white"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy" lang="EN-US">Yes Chip,</span><span style="font-size:9.0pt; font-family:"Helvetica",sans-serif" lang="EN-US"></span></p>
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<p class="MsoNormal" style="background:white"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy" lang="EN-US">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" lang="EN-US"></span></p>
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<p class="MsoNormal" style="background:white"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy" lang="EN-US">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" lang="EN-US"></span></p>
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<p class="MsoNormal" style="background:white"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy" lang="EN-US">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" lang="EN-US"></span></p>
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<p class="MsoNormal" style="background:white"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy" lang="EN-US">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" lang="EN-US"></span></p>
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<p class="MsoNormal" style="background:white"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy" lang="EN-US">I hope that helps make things clearer.</span><span style="font-size:9.0pt; font-family:"Helvetica",sans-serif" lang="EN-US"></span></p>
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<p class="MsoNormal" style="background:white"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy" lang="EN-US">Best regards,</span><span style="font-size:9.0pt; font-family:"Helvetica",sans-serif" lang="EN-US"></span></p>
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<p class="MsoNormal" style="background:white"><span style="font-size:10.0pt; font-family:"Arial",sans-serif; color:navy" lang="EN-US">Grahame</span><span style="font-size:9.0pt; font-family:"Helvetica",sans-serif" lang="EN-US"></span></p>
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