Bob, these three particles create a deuteron after all of the excess mass energy has been emitted as photons. The neutrino has very little energy because very little remains when the d forms. The creation process is unique to lenr and applies to all the isotopes of hydrogen, at least that is my model. if lenr is to be explained, you need to stop thinking in conventional terms. This is a new kind of nuclear process.
Ed Storms Sent from my iPad > On Feb 12, 2014, at 3:00 PM, "Bob Cook" <frobertc...@hotmail.com> wrote: > > Jones--Bob Cook Here-- > > Can you show how the p-e-p reaction as you understand it conserves spin? > > I would think that the newly fused particle, whatever it is, would have 1/2 > or 3/2 spin--I do not know. > > If a positron is emitted, its spin would be -1/2 I think. That would make > the new particle have 0 or 1 spin. > > The reaction of the positron and electron give photons with 0 spin. > > Bob > > > . > > -----Original Message----- From: Jones Beene tt > Sent: Wednesday, February 12, 2014 1:10 PM > To: vortex-l@eskimo.com > Subject: RE: [Vo]:a note from Dr. Stoyan Sargoytchev > > > > -----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 >
