No The traped photon scalar potential, could be described as
j_l(r w / c) Ylm(e) exp(iwt) if at the orbitsphere r w / c equal 0, then he adds a source terns of the form C Ylm(e) exp(iwt) to the scalar field equation and the outer part of the orbitsphere has zero scalar potential field (the static parts also cancelling) Now really what you have in Mills is Re(Ylm(e)exp(iwt) but that means that this photon field inside the orbitsphere is a standing wave. The nucleus is a more complex setup of EM + boundary conditions. Not sure that you could rule out a spherical wave. Who knows, but the fact is that the missing radiation is a mystery and I recon that in a cold fusion event in a solid state, it sure is many orders of magnitude more spherical symmetric than hot fusion, that's sounds like a clue to me. Also forget Mills, use QM if you don't like it. My argument is based in sound physical principles that is independent of math details. Of cause a spherical wave may be hard to melt and possibly un-physical. But we are here to fart ideas to each other. Smell it and tell me what you don't like or like about it - the mills versus QM, is really of less importance here Regards Stefan On Sat, Oct 10, 2015 at 5:25 AM, Eric Walker <eric.wal...@gmail.com> wrote: > Hi, > > On Fri, Oct 9, 2015 at 11:37 AM, Stefan Israelsson Tampe < > stefan.ita...@gmail.com> wrote: > > j_l(|w|/c r)Ylm(e)exp(iwt), with e the spherical part of x, and r the >> radial part. > > > Here you are using spherical harmonics -- Ylm(e). These are implicitly > disavowed by Mills, who offers instead the orbitsphere. You have suggested > in the past that there could be a duality that will resolve this > conundrum. I hope you will continue to give thought to the matter. > > Eric > >
