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 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]
<<attachment: winmail.dat>>
