<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 Albrecht,</div><div class=""><br class=""></div><div class=""> Thanks for confirming my calculations to your own satisfaction.</div><div class=""><br class=""></div><div class=""> Your derivation of the electron's inertia, which you claim relies on spatial extension within the electron, is unfortunately not relevant to the electron since there is no experimental evidence for separate parts of the electron. Perhaps your model will be found to be relevant to the electron some time in the future.</div><div class=""><br class=""></div><div class=""> In the meantime, it may be possible that an explanation for the electron's inertia can be found, based on an electron being composed of a single circulating spin 1/2 charged photon. Here’s a brief summary of my approach for your and others’ convenience. While my work is still a work in progress, I would like to know of any objections you see in it. The fact that the source of the centripetal force producing the internal circulation of the charged photon is currently unknown should not be an objection to the model as such. </div><div class=""><br class=""></div><div class=""><div class="">The electron’s relativistic energy-momentum equation is</div><div class=""><br class=""></div><div class="">E^2 = p^2 c^2 + m^2 c^4</div><div class=""><br class=""></div><div class=""> This can be rewritten as </div><div class=""><br class=""></div><div class="">(mc)^2 c^2 = E^2 - (pc)^2</div><div class=""><br class=""></div><div class="">So mc is just as invariant for the electron as the electron’s invariant mass m. But mc is a momentum. How should mc be interpreted? In my spin 1/2 charged photon model of the electron, mc is the circulating transverse component of the charged photon’s momentum, whose circulating total momentum is P = gamma mc, and whose energy E=gamma mc^2 is also the electron’s total energy E, and p= gamma mv is the longitudinal component of the charged photon’s total momentum, which is the same as the electron’s linear momentum p=gamma mv.</div><div class="">So since E=Pc in the charged photon model, the above equation can be rewritten as</div><div class=""><br class=""></div><div class="">(mc)^2 c^2 = (Pc)^2 - (pc)^2 </div><div class=""><br class=""></div><div class="">Dividing all terms by c^2 gives</div><div class=""><br class=""></div><div class="">(mc)^2 = P^2 - p^2 </div><div class=""><br class=""></div><div class="">or </div><div class=""><br class=""></div><div class="">(mc)^2 = (gamma mc)^2 - (gamma mv)^2 </div><div class=""><br class=""></div><div class="">So mc can be interpreted as the magnitude of the electron’s invariant internal momentum. It is the circulation of this internal charged photon momentum mc at a radius R=hbar/2mc that generates the electron’s spin hbar/2. It also generates the electron's inertial mass m, as described in my short article <a href="https://www.academia.edu/19652036/The_Origin_of_the_Electrons_Inertia" class="">https://www.academia.edu/19652036/The_Origin_of_the_Electrons_Inertia</a> .</div><div class=""><br class=""></div><div class="">See Figure 1 below from my spin 1/2 charged photon article at <a href="https://www.academia.edu/15686831/Electrons_are_spin_1_2_charged_photons_generating_the_de_Broglie_wavelength" class="">https://www.academia.edu/15686831/Electrons_are_spin_1_2_charged_photons_generating_the_de_Broglie_wavelength</a> .</div><div class=""><br class=""></div><div class="">In can be seen that mc is the invariant internal circulating momentum of the circulating spin 1/2 charged photon modeling the relativistic electron, where P=gamma mc is the charged photon’s total momentum and p=gamma mv is the charged photon’s longitudinal momentum.</div><div class=""><br class=""></div><div class=""> Richard</div><div class=""><br class=""></div><div class=""></div></div></body></html>