From: H Veeder (this also answers Robin's more recent posting)
>> The most elegant answer begins with the obvious assertion that there are no gammas ab initio, which means that no reaction of the kind which your theory proposes can be valid because gammas are expected. > RvS: Actually not only would I not expect to detect any gammas from a p-e-p reaction, I wouldn't expect to detect any energy at all. That's because the energy of the p-e-p reaction is normally carried away by the neutrino, which is almost undetectable. JB: the p+p reaction produces a positron, which annihilates with an electron producing 2 gammas. The net energy is over 1 MeV and easily detectable. Electron capture is real, but seldom by a proton at low energy. There is a real reaction in physics, but the ratio of that to p+p is 400:1 . so we have the insurmountable problem of exclusivity (see below). HV: The process of p-e-p fusion is suppose to be different from the process of p-p fusion. The outcome may be the same, but the processes differ. JB: Again, this is a very rare reaction - and my contention about it is twofold 1) there is no robust reaction in the real world where protons go directly to a deuteron without first forming a neutron, and that first step is energetically impossible, so the rarity of this p-e-p reaction is ingrained and systemic. 2) Therefore . even if there were such a reaction in LENR, at ten or even 100 times greater probability than the known p+p version, consider the obvious problem of exclusivity. Either way it does NOT happen in practice since we know there are no gammas ! Consider exclusivity. For the sake of argument - even if there are found to be two possible proton reactions, and one reaction is "supposed to be different" from the known solar reaction, but the outcome is the same except for the gamma - the problem always comes back to one of perfect exclusivity. Exclusivity is the logical fallacy that cannot be overcome. When a gamma reaction is known to happen with the same reactant, how can that reaction be excluded from happening, in a new scenario when both reactions are given enough energy to overcome the fusion threshold? Especially if one (the desired reaction) is much rarer than the other. Simplest answer: the known reaction cannot be excluded from happening, when the energy threshold is met - and there will be gammas even if the hypothetical p-e-p reaction has none by itself. ERGO. We really have no realistic option in framing a proper LENR theory - other than to find a gainful reaction which NEVER produces gammas nor indicia which are not in evidence (bremsstrahlung ). UV or soft x-rays are ok but no gammas Jones BTW - take an electron and proton at rest, that system has a mass of 0.511 + 938.272 = 938.8 MeV/c^2. That is the total mass available to that system. It cannot increase above that level unless substantial energy comes from outside the system. A neutron has a mass of 939.6 MeV/c^2. So, to make a neutron from an electron and a proton, the extra 782 keV has to come from outside the electron-proton system. It cannot come from the acceleration of the particles toward each other by their own attraction. One simply MUST make the neutron first - even if the deuteron, the end product of p+n does have a usable mass deficit. People who should know better are in denial about the rarity of p-e-p ! Let's get over it and move on. P-e-p is dead-in-the-water for adequately explaining the Rossi effect.
