<html><head><meta http-equiv="Content-Type" content="text/html charset=windows-1252"></head><body style="word-wrap: break-word; -webkit-nbsp-mode: space; -webkit-line-break: after-white-space;" class=""><div class="">Hello John,</div><div class=""><br class=""></div><div class=""> Thank you for your extended comments.</div><div class=""><br class=""></div><div class=""> First, I did not at all say that Einstein did not use insight, intuition and imagination. Obviously he did to a huge extent. “ <span style="color: red; font-size: 10pt; font-family: Cambria;" class="">You need to do what you were accusing Einstein of not doing! Intuition, insight and imagination!</span> “. I said that he could have probably increased his intuition (why does this idea offend some people?—Einstein was human too) through a systematic practice. Yogis call this practical approach intuitional science (though it might be more accurately called intuitional engineering since it refers to a practical, applied, systematic methodology). It doesn’t take the place of logic and reason. A yogi once said “Even if a small child says something logical it should be accepted. And even if the lotus-born Brahma’ (the mythological creator of the universe) says something illogical, it should be rejected like a straw.” </div><div class=""><br class=""></div><div class=""> Anyway, it could take me years (if ever) to develop enough intuition, insight and imagination to understand your recent electron model and its new high-dimensional algebra which is currently Greek to me. But I am eagerly waiting to hear of an endorsement of your new electron model from one or more highly-qualified independent physicists who DO understand all the math with all its subtleties. </div><div class=""><br class=""></div><div class=""> It the meantime, I hope you won’t continue to misunderstand my relatively simple electron model. You wrote: “ <span style="color: red; font-size: 10pt; font-family: Cambria;" class="">You cannot claim to get charge out if you put it in! </span><span style="color: red; font-size: 10pt; font-family: Cambria;" class="">Also – I have said this before and will not change my mind – you cannot put it in and stay with a massless photon. You just can’t Do the maths!</span><span style="color: red; font-size: 10pt; font-family: Cambria;" class=""> </span> “. But the circulating charged photon that I have proposed to model the electron is NOT massless. In my sentence just before your comment, I wrote : "<span style="font-size: 10pt; font-family: Cambria;" class="">the circulating double-looping photon is itself electrically charged and also has a rest mass of 0.511 MeV/c^2 and a spin of 1/2 hbar.</span> “ . So I am NOT AT ALL proposing that the charged spin 1/2 photon is massless. Its rest mass IS the electron’s rest mass, as its charge and spin ARE the electron’s charge and spin. How could this be otherwise if a circulating charged photon is to correctly model an electron? That would not be logical. Your new electron model also has associated electric charge -e, spin 1/2 and the electron's rest mass that it gets from its p-vot and its various other vots, if I may speak very simplistically and I’m sure imprecisely. </div><div class=""><br class=""></div><div class=""> Richard</div><div class=""> </div><br class=""><div><blockquote type="cite" class=""><div class="">On Nov 23, 2015, at 10:15 PM, John Williamson <<a href="mailto:John.Williamson@glasgow.ac.uk" class="">John.Williamson@glasgow.ac.uk</a>> wrote:</div><br class="Apple-interchange-newline"><div class=""><div style="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; direction: ltr; font-family: Tahoma; font-size: 10pt;" class=""><p class="MsoNormal" style="margin: 0cm 0cm 12pt; font-size: 12pt; font-family: Cambria;"><span style="font-family: 'Times New Roman';" class="">Hello Chip and Richard,</span></p><div style="margin: 0cm 0cm 12pt; font-size: 12pt; font-family: Cambria;" class=""><span style="font-family: 'Times New Roman';" class=""> </span><br class="webkit-block-placeholder"></div><p class="MsoNormal" style="margin: 0cm 0cm 12pt; font-size: 12pt; font-family: Cambria;"><span style="font-family: 'Times New Roman';" class="">I had been meaning to add to this post for some time, but did not find a free moment till now.</span></p><p class="MsoNormal" style="margin: 0cm 0cm 12pt; font-size: 12pt; font-family: Cambria;"><span style="font-family: 'Times New Roman';" class="">Will comment below, first on Chip’s post, then on Richard’s. This is also relevant to John Hodge's recent post on the nature of charge.<br class=""></span></p><p class="MsoNormal" style="margin: 0cm 0cm 12pt; font-size: 12pt; font-family: Cambria;"><span style="font-family: 'Times New Roman';" class="">Feel like going in red this morning ….</span></p><div style="margin: 0cm 0cm 12pt; font-size: 12pt; font-family: Cambria;" class=""><span style="font-family: 'Times New Roman';" class=""> </span><br class="webkit-block-placeholder"></div><p class="MsoNormal" style="margin: 0cm 0cm 12pt; font-size: 12pt; font-family: Cambria;"><span style="font-size: 10pt; font-family: Tahoma;" class=""><span class=""> </span>of comments from what a model…</span><span style="font-family: 'Times New Roman';" class=""></span></p><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: Cambria;" class=""><span style="font-size: 10pt; font-family: Times;" class="">Hi Richard</span></div><p class="MsoNormal" style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: Cambria;"><span style="font-size: 10pt; font-family: Times;" class=""> </span></p><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: Cambria;" class=""><span style="font-size: 10pt; font-family: Times;" class="">Correct me if I am wrong here. It seems that there is not a requirement that the electron actually be a sphere, but only that its scattering characteristics are the same as that of a sphere. Do you think this statement is correct?</span></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: Cambria;" class=""><span style="font-size: 10pt; font-family: Times; color: red;" class="">Yes and no. What is known is that the scattering is sphere-like – in that there is no “structure function” for the electron. This means, as I have said many times before, that the scattering is consistent with it being a SINGLE particle, with a spherical – inverse square law of scattering.</span></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: Cambria;" class=""><span style="font-size: 10pt; font-family: Times; color: red;" class="">Saying the electron must “be a sphere” anyway begs the question – what<span class=""> <span class="Apple-converted-space"> </span></span>kind of sphere? Is it a 3-sphere in 3-space? A four-sphere in 4D space? A sphere in the three components of the electric field (a bivector space)?<span class=""> <span class="Apple-converted-space"> </span></span>Something more complicated than any of these?</span></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: Cambria;" class=""><span style="font-size: 10pt; font-family: Times; color: red;" class="">I’m afraid, ladies and gentlemen, that the answer is the latter, though of the three specific static cases I think the third case comes closest. The electron, however, is certainly not static – it is very very dynamic.</span></div><p class="MsoNormal" style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: Cambria;"><span style="font-size: 10pt; font-family: Times;" class=""> </span></p><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: Cambria;" class=""><span style="font-size: 10pt; font-family: Times;" class="">Chip</span></div><p class="MsoNormal" style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: Cambria;"><span style="font-size: 10pt; font-family: Times;" class=""> </span></p><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: Cambria;" class=""><b class=""><span style="font-size: 11pt; font-family: Calibri;" class="">From:</span></b><span style="font-size: 11pt; font-family: Calibri;" 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>Thursday, November 19, 2015 7:46 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="">Cc:</b><span class="Apple-converted-space"> </span>Nick Bailey <<a href="mailto:nick@bailey-family.org.uk" style="color: purple; text-decoration: underline;" class="">nick@bailey-family.org.uk</a>>; David Williamson <<a href="mailto:david.williamson@ed.ac.uk" style="color: purple; text-decoration: underline;" class="">david.williamson@ed.ac.uk</a>>;<span class="Apple-converted-space"> </span><a href="mailto:pete@leathergoth.com" style="color: purple; text-decoration: underline;" class="">pete@leathergoth.com</a>; Mark, Martin van der <<a href="mailto:martin.van.der.mark@philips.com" style="color: purple; text-decoration: underline;" class="">martin.van.der.mark@philips.com</a>><br class=""><b class="">Subject:</b><span class="Apple-converted-space"> </span>Re: [General] Reply of comments from what a model…</span><span style="font-size: 10pt; font-family: Times;" class=""></span></div><p class="MsoNormal" style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: Cambria;"><span style="font-size: 10pt; font-family: Times;" class=""> </span></p><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: Cambria;" class=""><span style="font-size: 10pt; font-family: Times;" class="">Hello John D and Albrecht,</span></div><p class="MsoNormal" style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: Cambria;"><span style="font-size: 10pt; font-family: Times;" class=""> </span></p><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: Cambria;" class=""><span style="font-size: 10pt; font-family: Times;" class=""> We’re not quite there by merely replacing Albrecht’s two circulating massless particles by a double-looping photon. By doing this the radius of the circle drops from hbar/mc to hbar/2mc because the total loop length is still one Compton wavelength. A double loop of length 1 Compton wavelength h/mc has half the radius of a single loop and therefore (if the circulating photon carries charge -e moving at light speed) half the calculated magnetic moment of Albrecht’s model, i.e. 1/2 Bohr magneton. The loss in magnetic moment from Albrecht’s 2-particle model has to be made up in some other way. But this double-looping photon model of the electron has spin 1/2 hbar while Albrecht's two-particle model has spin 1 hbar. No argument about retarded light-speed forces between his 2 light-speed circling massless particles will bring the total spin of the two-particle system down to exactly 1/2 hbar while keeping its magnetic moment at 1 Bohr magneton. That would be like pulling a magical rabbit out of a hat which so far only Dirac with his equation has been able to do successfully (he wasn’t called a magician for nothing.) The Williamson - van der Mark 1997 electron model comes close with its proposed centrally located static electric charge -e inferred from their twisting double-looping uncharged photon’s inward pointing electric fields at the model’s equator.<span class="Apple-converted-space"> </span></span></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: Cambria;" class=""><span style="font-size: 10pt; font-family: Times; color: red;" class="">The WvdM model does get the magic rabbit right. Not only that it gets the QED first order correction to the magic rabbit right (about 1 part in a thousand bigger) – which the Dirac model does not do.</span></div><p class="MsoNormal" style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: Cambria;"><span style="font-size: 10pt; font-family: Times;" class=""> </span></p><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: Cambria;" class=""><span style="font-size: 10pt; font-family: Times;" class="">(But what happened to their double-looping photon's electric field at and near the model’s two poles?) .<span class="Apple-converted-space"> </span></span></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: Cambria;" class=""><span style="font-size: 10pt; font-family: Times; color: red;" class="">Richard, you are still thinking about a little photon bullet whizzing around in 3-space only. This is not good enough. You need to do what you were accusing Einstein of not doing! Intuition, insight and imagination!</span></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: Cambria;" class=""><span style="font-size: 10pt; font-family: Times; color: red;" class="">The original<span class=""> <span class="Apple-converted-space"> </span></span>1997 paper already explained the transport around the torus was not in space but in space-time. The rotations are not just in 3-space but in a higher-dimensional space. In three space one cannot have, simultaneously the two axes of “rotation” that are needed for the WvdM model. In 4-space one can. This is the “quantum bicycle” I keep trying to explain to you. A 4-spatial rotation is still (in my present view) too simple, but illustrates (one of the) salient points. Imagine a space x y z w. Now allow a rotation in the xy plane, with a simultaneous rotation in the zw plane. Now let the path traced by a point (x y z w) fill 4-space. Let the length of this path (x squared plus y squared plus z squared plus w squared) oscillate in phase with “rotations”. This is the program I implemented in the little java applet I circulated a few months ago.<span class="Apple-converted-space"> </span><span class=""> </span>What does one observe when one projects this “motion” onto 3-space? You can find lots of these projections on the web if you look. It is kind of difficult to do it in your head – but dead easy to implement it in a computer . Anyway, in one kind of projection one observes a sphere, in another a torus. For such flows, it is perfectly possible (even necessary) to have a spherical projection for the electric field, while having a toroidal form in a projection onto other spaces. Thinking in just 3D space severely limits ones imagination!</span></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: Cambria;" class=""><span style="font-size: 10pt; font-family: Times; color: red;" class="">Now the motion I’m envisioning nowadays is more complicated than merely 4-dimesional, as there are far more “planes” than just the six in 4-D space. The electron rotation has three rotation planes (at least!) Looking at the photon solution (eq 21) one rotation is a normal spatial plane (xy), the other in the “plane” formed from the scalar and the pseudoscalar. This latter pair are isomorphic to complex numbers. This means the photon “twist” is already in a 4-component space, just not that of x y z t, but that of scalar, pseudocalar, electric and magnetic field “space”. Now to get the electron solution, one takes that<span class=""> <span class="Apple-converted-space"> </span></span>already “4-dimensional” motion and lets it loop again “rotating” it in yet another plane in the even subset (of eight!) dimensions.<span class=""> <span class="Apple-converted-space"> </span></span>The resulting object is rotating in (at least) nine “dimensions” (eight modulated by “time”). What one observes is a projection of this. What is required by experiment is that the interaction part (the electric field part) is spherical, at least if one does not come within touching distance when direct field interference kicks in. At these distances the Pauli exclusion principle kicks in, as described in my 2012 paper at MENDEL.</span></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: Cambria;" class=""><span style="font-size: 10pt; font-family: Times;" class="">This model can’t convincingly explain how a sphere enclosing a double-looping uncharged photon can have a non-zero divergence of its electric field (indicating a non-zero enclosed electric charge) without violating Gauss’ law (the first Maxwell equation).</span></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: Cambria;" class=""><span style="font-size: 10pt; font-family: Times; color: red;" class="">This is only true if you take the electron to be constituted a massless photon (as you do).<span class=""> <span class="Apple-converted-space"> </span></span>Let me try, once again, to convince you.</span></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: Cambria;" class=""><span style="font-size: 10pt; font-family: Times; color: red;" class="">Look at Gauss’s law in the full set of equations in my paper.<span class=""> <span class="Apple-converted-space"> </span></span>This is equation 6. There is another term, as well as the electric field divergence (which is the DEFINITION of “charge”) corresponding to root-mass exchange.<span class=""> <span class="Apple-converted-space"> </span></span>This is the nature of charge in QED. The electric field divergence, in the new equations, is non zero if there is mass-energy exchange.<span class=""> <span class="Apple-converted-space"> </span></span>That is (part of) the root of charge. It is not the whole story – as photon exchange needs ALL eight (well at least seven) of the even terms to explain it properly. It does mean that Gauss’s law needs to be extended by allowing for mass-energy exchange though. This is anyway the case, if you think about it, in both QED and the inhomogenous Maxwell equations (where,in both, you put in the “charge by hand!).</span></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: Cambria;" class=""><span style="font-size: 10pt; font-family: Times; color: red;" class="">Given the state- of play of Martin and my model in 2015 there are now two ways to calculate the charge in the resulting model. The first is to use the curvature, and the calculated electric field, to get the charge in terms of Plancks’ constant (or vice versa). This is what Martin and I did in out 1997 paper. The other way is to integrate the cross-section of charge-charge interactions over the universe – which requires a knowledge of the number of charges in the universe and their distribution. This is harder. Both give values for the elementary charge within the right ballpark, however.</span></div><p class="MsoNormal" style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: Cambria;"><span style="font-size: 10pt; font-family: Times;" class=""> </span></p><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: Cambria;" class=""><span style="font-size: 10pt; font-family: Times;" class="">I think that in order to retain a viable double-looping photon model of the electron, one may have to bite the bullet and accept that the circulating double-looping photon is itself electrically charged and also has a rest mass of 0.511 MeV/c^2 and a spin of 1/2 hbar.</span></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: Cambria;" class=""><span style="font-size: 10pt; font-family: Times; color: red;" class="">Absolutely not! You cannot claim to get charge out if you put it in! Also – I have said this before and will not change my mind – you cannot put it in and stay with a massless photon. You just can’t Do the maths! Integrate the mass-energy in any one frame due to the charge alone and you will get a non-zero mass. This mass will be minimal where the field is radial – and will increase for any other frame. End of story. You can SAY you have a “charged massless photon”– but this does not make it consistent with reality! Sorry!<a name="_GoBack" class=""></a></span></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: Cambria;" class=""><span style="font-size: 10pt; font-family: Times; color: red;" class="">You can say (and be right) that you have a charged electron with rest mass (if this is what you mean) – but this is just what we have all been saying all along – so what is the difference?</span></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: Cambria;" class=""><span style="font-size: 10pt; font-family: Times;" class=""> By the way, Albrecht’s two circulating particles may each have no rest mass as he describes, but they certainly each carry 1/2 of 0.511 MeV of a resting electron's total energy. This strongly implies that they are two circulating photons (or gluons?) each having energy 1/2 x 0.511 MeV. This also gives his electron model a spin of 1 hbar.</span></div><p class="MsoNormal" style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: Cambria;"><span style="font-size: 10pt; font-family: Times;" class=""> </span></p><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: Cambria;" class=""><span style="font-size: 10pt; font-family: Times;" class=""> with best regards,</span></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: Cambria;" class=""><span style="font-size: 10pt; font-family: Times;" class=""> Richard</span></div><div style="margin: 0cm 0cm 0.0001pt; font-size: 12pt; font-family: Cambria;" class=""><span style="font-size: 10pt; font-family: Times; color: red;" class="">Regards, from John.</span></div></div><span style="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; 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