<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 Albert (and all),</div><div class=""><br class=""></div><div class=""> I think your fundamental objection that you mentioned earlier can be answered below.</div><div class=""><br class=""></div><div class=""> The left side of the big triangle in Figure 2 in my article is a purely mathematical unfolding of the path of the helical trajectory, to hopefully show more clearly the generation of de Broglie wavelengths from plane waves emitted by the actual charged photon moving along the helical trajectory. Nothing is actually moving off into space along this line.</div><div class=""><br class=""></div><div class=""> Consider an electron moving with velocity v horizontally along the helical axis. Since in Figure 2 in my article, cos (theta) = v/c , the corresponding velocity of the charged photon along the helical path is v/ cos(theta) = c , the speed of the charged photon, which we knew already because the helical trajectory was defined so that this is the case. In a short time T, the electron has moved a distance Delectron = vT horizontally and the photon has moved a distance Dphoton = Delectron/cos(theta) =vT/cos(theta) = cT along its helical trajectory. A plane wave front emitted from the photon at the distance Dphoton = cT along the photon’s helical path will intersect the base of the big triangle (the helical axis) at the distance along the base given by Dwavefront = Dphoton / cos(theta) = cT/ (v/c) = T * (c^2)/v which means the intersection point of the plane wave with the helical axis is moving with a speed c^2/v which is the de Broglie wave’s phase velocity. The length of the de Broglie wave itself as shown previously from Figure 2 is Ldb = Lambda-photon / cos(theta) = h/(gamma mc) / (v/c) = h/(gamma mv). So as the electron moves with velocity v along the z-axis, de Broglie waves of length h/(gamma mv) produced along the z-axis are moving with velocity c^2/v along the z-axis. The de Broglie waves created by the circulating charged photon will speed away from the electron (but more will be produced) to take their place, one de Broglie wave during each period of the circulating charged photon (corresponding to the moving electron). As mentioned previously, the period of the circulating charged photon is 1/f = 1/(gamma mc^2/h) = h/(gamma mc^2/). As the electron speeds up (v and gamma increase) the de Broglie wavelengths h/(gamma mv) are shorter and move more slowly, following the speed formula c^2/v .</div><div class=""><br class=""></div><div class=""></div></body></html>