Frederick Sparber writes, > OTOH, for two deuterons the (repulsive charge)/(n*quark-magnetic force) ratio is 2*Z/14 not counting the antineutrino in each.
> How would this work for three BEC helium nuclei with 3*Z/42? :-) Well, Fred, they look pretty doggone close, don't they? What you seem to be saying is - that in terms of the relative repulsive force which is felt in condensed matter at picometer distance, which when overcome will lead to fusion, the "net" force between three helium nuclei is actually no greater at all than what would be felt by two deuterons, no? One might further opine that if BEC-like cold fusion does occur in condensed matter, one might well be advised to use He rather than D2 because He is "always" a boson, while D2 is only bosonic during the time that its electron is not closely bound, which one can assume has some positive value ...that is, assuming that one can load He in such a way (ion implantation) that 3 atoms are usually present in a single vacancy. Not to mention costs less...and if your experiment does not work, you can inhale it and sing like Roy Orbison, or else fill party balloons.... Jones Overheard at the Palladium, a former CF lab, now a Karaoke bar with a helium tank... tastes great!! no, less filling tastes great!! no, less filling tastes great!! no, less filling tastes great!! no, less filling
