Jones Beene <[EMAIL PROTECTED]> wrote:
>
> 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?
>
I would say so.
>
> 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?
>
Yes, but if the quarks (current loops or disks) are "Hertzian Dipoles"
oscillating at 1.0e25 Hz
the magnetic force should be 1/R^2 making it greater (attracting) than the
repulsive coulomb force
if the (solenoids-like) alignment is favorable. OTOH, like mini-bar magnets
under BEC conditions they
should align to attract as opposed to Hot Fusion conditions that randomizes
the alignment.
>
> 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.
>
Okay.
>
> 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....
>
Or Wayne Newton? :-)
>
Frederick
> Jones
