Come to think of it, this pion reaction is a mind bender. This Mizuno reaction is the exact opposite of the proton-proton chain reaction. That is the fusion reaction by which stars convert hydrogen to helium. If the proton-proton chain reaction produces positive energy, then its opposite fission reaction must need the same energy supplement to occur.
If the Mizuno reaction is occurring, the energy needed for it to occur must be coming from the vacuum, fed by the magnetic field that connects the location of vacuum activity to the nucleus. This Mizuno experiment is such a pure, seemingly uncomplicated, and straightforward experiment, the basic reaction is self-evident, yet it is unbelievable by all conventional standards. They say that the data never lies; but wow, does LENR really get all or most of its energy from the vacuum. On Sat, Mar 29, 2014 at 1:14 PM, Axil Axil <janap...@gmail.com> wrote: > I favor the idea that a strong magnetic field can catalyze pion virtual > particles from the energy borrowed from the vacuum that can disrupt the > nucleus of the atom along the path of that magnetic field. > > > If a plus pion transmutes a neutron into a proton using borrowed energy > from the vacuum (135 MeV), no angular momentum is carried into the > reaction. Furthermore, all energy issues would be resolved through > instantaneous decay of the neutron as transformed at that instant by the > plus pion. That is, no 14 minute neutron decay time would occur. > > No electron and neutrino would be produced in the reaction since the > neutron to proton transmutation process is a run of the mill nuclear one > and not a radioactive one. The energy of the reaction being generated is > simply the energy equivalent of the mass difference between the deuterium > atom and 2 protons. That energy would be transferred by the magnetic field > to the origin of the magnetic field and thermalized without loss in a dark > mode within a positive feedback loop. > > The spin of the D being one would be broken into 2 spin 1/2 protons thus > with spin having been conserved. > > > > > > On Sat, Mar 29, 2014 at 11:20 AM, Jones Beene <jone...@pacbell.net> wrote: > >> Part II >> >> When a free neutron decays to a proton, substantial energy is released as >> well as a neutrino - which carries away about 40% of the net energy >> undetected. That is the main problem to overcome in framing a putative >> exothermic deuterium reaction in place of the endotherm which would >> normally >> appear. There is a valid QM rationalization for this, but the probability >> of >> it happening is unknown. >> >> Outside the nucleus, free neutrons are unstable and have a mean lifetime >> of >> about 15 minutes. They beta decay with the emission of an electron and >> electron antineutrino leaving a fairly cold proton. The decay energy for >> this process is up to 0.78 MeV for the electron, but is highly variable- >> unlike almost any other nuclear reaction. The energy of the unseen >> neutrino >> which is emitted is about 500 keV on average - which explanation resolves >> problems of conservation of spin and the lower net energy which is >> sometimes >> seen in experiment. >> >> The variability of energy release is hard to reconcile without a "kludge" >> of >> some kind - which is the neutrino. The reality of the neutrino in general >> is >> not in question here, but its application to a related reaction is in >> question, since it may not be required when the need is obviated. >> >> The free neutron mass is larger than that of a proton: 939.565378 MeV >> compared to 938.272046 MeV. The difference is ~1.3 MeV. Since the apparent >> energy release from neutron decay is occasionally nearly the entire value >> of >> the theoretical mass difference, we must ask: is the neutrino really >> necessary in a D+D collision, or any other without "allowed spin" >> problems, >> or is a relic of trying something else which has taken on a life of its >> own? >> When two neutrons decay together immediately on the impact of two >> deuterons >> which do not have enough momentum to fuse, the collision can be a mini QCD >> version of "quark soup" that seldom overcomes the barrier for fusion to >> helium, but is nevertheless energetic. Moreover there is no allowed spin >> problem. >> >> Consider the spins of the electron and antineutrino with a net spin of >> zero. >> This is a "Fermi decay" since the electron and antineutrino take no spin >> away, and the nuclear spin cannot change. The other possibility allowed by >> QM is that spins combine into a net spin of one: "Gamow-Teller decay." The >> angular momentum can change by up to one unit in an allowed "double beta" >> decay, which is the closest analogy. Consequently, there is a distinct >> possibility for spin issues to be resolved in the context of two >> inseparable >> reactions involving deuterons, but without neutrino emission. >> >> There is another issue - the extended half-life of free neutrons - which >> means that decay energy is not normally available instantaneously, to >> "lend" >> in the sense of quantum mechanics. This is where QM enters the picture in >> two different ways. The mass of the deuteron is 1875.613 MeV. The mass of >> a >> free neutron plus a free proton is 1877.8374 - thus about 2.2 MeV would be >> required (to be supplied via kinetic energy) in order to split the >> deuteron >> - without QM. The net deficit of this reaction is somewhere around ~900 >> keV >> if the neutrino is avoided. So far, even assuming a time reversed >> borrowing, >> we are still at endotherm unless the same initial kinetic energy provides >> two identical reactions. Voila! ... then there is net gain to the extent >> neutrino release is avoided. >> >> An apparent endotherm is the only reason why no one ever imagined >> Oppenheimer Philips as being relevant before now. It looks endothermic, >> without Heisenberg uncertainty - and even more so without neutrino >> suppression. However, one can surmise that when two deuterons approach >> each >> other so that both undergo the OP splitting reaction instantaneously as a >> result of the single impact, then the same 2.2 MeV of kinetic energy >> results >> in both reactions. This is an implication of Heisenberg. A net energy >> release of 2.6 MeV is then seen (from two instantaneous neutron decays >> without neutrinos). Most of the threshold energy can be borrowed. The two >> neutrons have decayed to protons instantly, instead of with an extended >> half-life and we have an allowed spin state without neutrino release. >> >> Thus the net reaction gain is 300-400 keV imparted to two electrons. The >> stretch of the imagination is that the same kinetic energy can split both >> atoms at exactly the same time, invoking quantum uncertainty. Thus, using >> borrowed energy from the net reaction - with neutrino emission suppressed >> we >> now have a net gainful reaction. Admittedly, this is a stretch, but isn't >> everything in QM, especially when first invoked ? >> >> The reality of this or any such QM explanation for an experimental result >> is >> dependent on the accuracy of Mizuno's mass spectroscopy. If Mizuno is >> correct, this is a defensible first step to consider towards a viable >> answer >> to the finding (of twice the quantity of gas in the ash of the reaction). >> Can anyone propose another defensible hypothesis for gain, giving benefit >> of >> doubt to Mizuno, which can support these findings? >> >> There is the possibility of the fractional hydride of deuterium, ala >> Randell >> Mills, which would be a stable negative ion of mass 2. However, a reactor >> full of deuterino hydride [f/D-] would have the same electrostatic >> repulsion >> problem as a reactor full of D+. Thus, H2 seems like the most likely >> species >> for a mass two gas which has doubled in quantity. Mills supporters might >> say >> that what we have is ionic bonds between fractional deuterium hydride and >> the positive deuteron ion. That avenue cannot be immediately rejected, and >> should be explored as a valid alternative. >> >> One other possibility to get there - is that D reacts with Ni, and two >> protons are released from the reaction for a net gain of energy plus >> double >> the gas quantity, as reported. This implies a transmutation of nickel to >> iron. But there is no known reaction to support this conclusion, and no >> showing of transmuted iron. Plus, nickel and iron are the two most stable >> of >> all nuclei in the periodic table and a low energy transmutation is highly >> improbable. >> >> In conclusion, there is a version of a known double beta decay reaction to >> support the conclusion of an energetic splitting of two deuterons into >> four >> protons. There is even the expectation of a neutrino-less version, which >> adds about 1 MeV to the bottom line - and has some high powered support at >> SLAC. >> >> >> http://www.symmetrymagazine.org/article/august-2013/neutrinoless-double-beta >> -decay<http://www.symmetrymagazine.org/article/august-2013/neutrinoless-double-beta-decay> >> >> Note that in normal beta decay about 500 keV+ is lost to the neutrino. >> That >> is the crux of the situation. Can neutrino retention be rationalized via >> QM >> principles in order to make an seemingly endothermic reaction gainful? >> >> Not exactly easy to do, or we would not be having this conversation - yet, >> the deuteron is weakly bound and nickel is strongly bound, so there is a >> real probability that the bulk of the Mizuno reaction involves only >> deuteron >> collisions for gain, with nickel as the nanomagnetic catalyst and protons >> as >> the ash. >> >> _____________________________________________ >> Part I was posted separately under this same subject >> heading >> >> The recent Mizuno (Yoshino) presentation at the MIT >> colloquium and the surprising implication of finding about twice the >> quantity of hydrogen appearing as ash from deuterium reactions (as the >> starting gas) after a month long run - has been the inspiration for the >> following early stage hypothesis. This is a revision to focus on >> nano-magnetics and the SPP contribution. [snip] >> >> >
