I've heard Rossi and some others happy about some signal around 511Kev (e+ anihilation)... to be confirmed.
note that DGT claim gamme in 30-500keV... compatible with 511keV divided (is it possible? ) however if much energy is cared by e+, and annihilation, should not there be much more gamma than observed, at a point of being dangerous ? Just a naive question... if no gamma nor neutrons is produced at noticeable quantity, does it mean that most energy is transmitted by some charged particles, that don't annihilate ? why not alpha, e-, p+, heavy ions, all with important kinetic energy, which is dispersed in many quanta in the lattice by many electromagnetic interaction question to physicist: if an e-, or a p+, an alpha is thrown out of the reaction zone with say as much as 24MeV kinetic energy (or twice 12MeV), how is the kinetic energy dissipated ? something like Cerenkov ? are there many gamma produced? 2014-02-12 23:21 GMT+01:00 H Veeder <hveeder...@gmail.com>: > > > > On Wed, Feb 12, 2014 at 4:10 PM, Jones Beene <jone...@pacbell.net> wrote: > >> >> >> -----Original Message----- >> From: mix...@bigpond.com >> >> >The most elegant answer begins with the obvious assertion that there are >> no >> gammas ab initio, which means that no reaction of the kind which your >> theory >> proposes can be valid because gammas are expected. >> >> Actually not only would I not expect to detect any gammas from a p-e-p >> reaction, I wouldn't expect to detect any energy at all. That's because >> the >> energy of the p-e-p reaction is normally carried away by the neutrino, >> which >> is almost undetectable. >> >> Hi, >> >> Not so - the reaction produces a positron, which annihilates with an >> electron producing 2 gammas. They net energy is over 1 MeV and easily >> detectable. >> >> Jones >> > > > The process of p-e-p fusion is suppose to be different from the process of > p-p fusion. > The outcome may be the same, but the processes differ. > > > Harry >
