This brings up the down side of LENR. LENR produce intense ionization produced by muons far from the reaction. Many experimenters have complained about this intense ionization which makes electronics useless.
On Thu, Jan 19, 2017 at 4:01 PM, Jones Beene <jone...@pacbell.net> wrote: > > You still are not making the correct and important distinctions from this > paper. > > This may sound pedantic but "decay" is not the same thing as > "annihilation." If is important to use the correct semantics here. > > See: https://en.wikipedia.org/wiki/Particle_decay > > 1) Mesons are derived from annihilation of the proton, NOT decay of > protons. > > 2) Mesons decay to muons. Muons decay to lighter leptons. > > 3) Protons do not decay. At least not in 10^29 years - far longer than the > age of the Universe > > 4) A laser pulse is required to produce the annihilation event in protons > - the weak force is not involved at this point. > > 4) A huge amount of energy is produced from annihilation, much more than > any decay event. > > 5) This energy is generally NOT USABLE as the muons disperse far from the > reactor. > > 6) To obtain usable energy, then actual fusion must be incorporated into > the system. > > 7) Fusion of deuterons is a secondary effect of muons, which catalyze > deuterons. > > 8) Without fusion the energy of the muon decay is essentially lost > hundreds of meters away. > > 7) Because deuterium fusion in this case produces charged particles of >3 > MeV - that energy can be captured and not lost. There are few gammas. > > Thus we have a catch-22 scenario. The extreme energy of proton > annihilation to mesons and muons is difficult to capture, and thus > breakeven or net gain requires a secondary reaction - fusion - using > deuterium. As of now, Holmlid has not shown a way to reach breakeven > without deuterium fusion being the primary source of USABLE energy. > > On 1/19/2017 12:00 PM, Axil Axil wrote: > > 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 <jone...@pacbell.net> 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 >> > > >
