<html><head><meta http-equiv="Content-Type" content="text/html charset=utf-8"></head><body style="word-wrap: break-word; -webkit-nbsp-mode: space; -webkit-line-break: after-white-space;" class=""><div class="">Hello Vivian,</div><div class=""><br class=""></div><div class=""> Thank you for your detailed reply. It’s always better for one’s written mistakes to be pointed out by a well-wisher. An arXiv article <a href="http://arxiv.org/abs/physics/0512265" class="">http://arxiv.org/abs/physics/0512265</a> by Qiu-Hong Hu on “The nature of the electron"</div>
<div class="">made the same mistake (on bottom of p.13) in calculating the magnetic moment of a double-looped electron. I emailed him twice pointing out his error but he never replied (so I’m not sure he received my emails), but his article was never corrected on arXiv (I just checked.)</div><div class=""><br class=""></div><div class=""> Here’s an error in your article that you can correct quickly: At the bottom of page 13 in your article, the vertical leg of the right triangle is labeled KE = pc = h Fke . But the kinetic energy KE of a relativistic electron is not equal to pc= (gamma mv)c . The relativistic kinetic energy KE of a particle is KE = (gamma-1)mc^2 which is clearly not equal to pc = (gamma mv)c although both terms go to zero as v approaches zero. Actually the vertical leg of your triangle is correctly labeled as pc but not as KE, according the the relativistic energy-momentum equation E^2 = (pc)^2 + (mc^2)^2 . It is also correct that E = gamma mc^2 = mc^2 + KE = mc^2 + (gamma-1)mc^2 . But this is a linear and not squared sum since energies add linearly as scalar quantities. </div><div class=""><br class=""></div><div class=""> Now, it turns out that there is some usefulness in equating KE with h Fke as you did, although KE = h Fke should not be on the vertical leg of your triangle. On the second page of my article at <a href="https://www.academia.edu/10235164/The_Charged-Photon_Model_of_the_Electron_Fits_the_Schrödinger_Equation" class="">https://www.academia.edu/10235164/The_Charged-Photon_Model_of_the_Electron_Fits_the_Schrödinger_Equation</a> (attached below), I show that the electron’s kinetic energy KE can be expressed as hbar w (where w means angular velocity omega), or as you expressed on the vertical leg of your triangle: KE = h Fke , where Fke = w/2pi. This result is derived from the earlier expression in my article that the total energy E of the circulating charged photon is given by E=h w(total) = h (wo + w) where h wo corresponds to mc^2 of a resting electron and h w corresponds to the electron’s KE. Since E=mc^2 + KE = h wo + h w and since mc^2 = h wo, it follows by subtraction that KE = h w which corresponds to your formula KE = h Fke . In the non-relativistic limit (where Schrodinger’s equation is used) we have KE=(p^2)/2m . So for a free non-relativistic electron we have (p^2)/2m = h w . And for a non-relativistic electron in an electric potential well V(x) we have (p^2)/2m + V(x) = h w . You can see how this leads to the time dependent Schrodinger equation: </div><div class=""> -hbar^2/2m d^2 PSI (x,t) /dt^2 + V(x) PHI (x,t) = i hbar d PSI(x,t)/ dt where for a free electron ( V(x)=0 ) , PSI =Ae^i(kx-wt) where p=hbar k and KE=hbar w</div><div class=""> </div><div class=""> So your Fke = w/2pi corresponds to the w in the time-dependent Schrodinger equation. This correspondence is in my opinion is why Schrodinger’s time dependent (and also time independent) equation works — its basic formula KE + V = hw comes from the circulating spin-1/2 charged photon model of the electron with its relativistic momentum equation P^2 = (mc)^2 + p^2 where P=E/c , which corresponds to the electron’s relativistic energy-momentum equation E^2 = p^2 c^2 = m^2 c^4 .</div><div class=""><br class=""></div><div class=""> The other significant mistake in your paper was in your electron radius derivation, by treating the time interval t as the same in two different reference frames, although it should have been t’=gamma t (t being the proper time), where gamma comes into the derivation of time dilation because the speed of light c is (taken to be) constant in all inertia frames. You choose to not correct this mistake, claiming that you will go by experimental results. But you are not the only one predicting (from your incorrect calculation) that the transverse radius of an electron decreases as 1/gamma with the electron’s increasing velocity. John Williamson claimed the same relationship 1/gamma relationship based on general energy considerations for a moving electron. In my spin-1/2 charged-photon model of the electron, where the radius of the charged photon is directly proportional to the photon’s wavelength (R=L/4pi) and therefore inversely proportional to its energy, also gives the same 1/gamma relationship the change of the transverse electron radius with increasing electron speed. </div><div class=""><br class=""></div><div class=""> If you will not believe me, I request that someone like John W or Martin (or anyone knowing basic relativistic time dilation calculations) take a few minutes to check your time-electron radius calculation in your article and give his opinion to all of us. If they concur with my result and you still wish to insist that only experiment will tell if your radius calculation is correct, then I have nothing more to say on this.</div><div class=""><br class=""></div><div class=""> with best regards,</div><div class=""> Richard</div><div class=""><br class=""></div><div class=""><br class=""></div><div class=""><br class=""></div><div class=""><br class=""></div><div class=""></div></body></html>