Holmlid states as follows: The state *s* = 1 may lead to a fast nuclear reaction. It is suggested that this involves two nucleons, probably two protons. The first particles formed and observed [16 <http://journals.plos.org/plosone/article?id=10.1371/journal.pone.0169895#pone.0169895.ref016> ,17 <http://journals.plos.org/plosone/article?id=10.1371/journal.pone.0169895#pone.0169895.ref017>] are kaons, both neutral and charged, and also pions. From the six quarks in the two protons, three kaons can be formed in the interaction. Two protons correspond to a mass of 1.88 GeV while three kaons correspond to 1.49 GeV. Thus, the transition 2 p → 3 K is downhill in internal energy and releases 390 MeV. If pions are formed directly, the energy release may be even larger. The kaons formed decay normally in various processes to charged pions and muons. In the present experiments, the decay of kaons and pions is observed directly normally through their decay to muons, while the muons leave the chamber before they decay due to their easier penetration and much longer lifetime.
Holmlid recognized that the DECAY of protons is where the mesons come from. This decay is a weak force reaction in which a huge amount of energy is produced...(1.88 GeV while three kaons correspond to 1.49 GeV). Deuterium has nothing to do with proton decay. The protium nanoparticle can produce proton decay just as well as deuterium. The protium nanoparticle will still produce the 1,88 GeV as well as the deuterium nanoparticle. Fusion is just as secondary side issue. On Thu, Jan 19, 2017 at 2:29 PM, Jones Beene <[email protected]> wrote: > Axil Axil wrote: > > The first reaction to occur is meson production which as nothing to do > with fusion: > > > Well, that is partially true - mesons come first after the laser pulse. No > one cares, since mesons have incredibly short lifetimes. > > The main point is that mesons very quickly into muons. *Muons catalyze > fusion in deuterium.* > > Muon catalyzed fusion has been known for 75 years. It would be next to > impossible to avoid fusion when muons and deuterons are both present. > > The bottom line is this: if there is to be net gain, deuterium must be > used because fusion provides the usable gain - not mesons or muons which > decay too far away to provide gain. > > Jones >
