In reply to Jürg Wyttenbach's message of Wed, 12 Jun 2019 03:24:41 +0200: Hi, [snip] >Regarding deep orbits: > >There is absolutely no physical solution for the forces for any QM based >model for deep orbits. The basic rules of any physical model that >includes mass are given by the de Broglie radius. Any violation of the >coulomb mass/EM-mass relation needs an additional explanation by a new >physical concept, that has never been given by anybody that modeled deep >orbits. > >E.g. a deep orbit of 400keV means that the electron mass classically >should increase to a manyfold value of 400keV. But there is no mechanism >to increase the classic central force if we do not include magnetic >central forces. But these forces are not covered by QM and need a >different treatment based on rotations only!
Mills uses a "pseudo" charge increase to supply an increased central force. I have taken a somewhat different approach, allowing the De Broglie wave to wrap around multiple times in three dimensions before reconnecting. That results in no increase in central force needed. See http://rvanspaa.freehostia.com/relativistic-both.pdf (This is probably due to be revamped, but frankly I can't be bothered at the moment. ;) A consequence of this is that the angular momentum of the electron takes on fractional values which in turn explains why the "ground state" of the hydrogen atom doesn't radiate, i.e. the difference in angular momentum between any two of the sub-orbitals in insufficient to provide the angular momentum required for a photon. In short no photon can be formed => no radiation => explains normal stability of hydrogen ground state atom. This doesn't prevent the lower states being accessed, provided that an alternate method (not based on photons) is available to share the angular momentum. This approach yields the same energy levels as Mills, but different radii. Mills' radii are proportional to 1/n whereas mine are proportional to 1/(n^2). QM assumes that the electron angular momentum is quantized, whereas I contend that this only appears to be so as a consequence of the fact that all photons have the same angular momentum. (For elliptically polarized photons, the angular momentum vector doesn't align with the trajectory, but has the same absolute magnitude.) > >Jürg [snip] Regards, Robin van Spaandonk local asymmetry = temporary success
