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
<mailto: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
