Bob,
I do not necessarily disagree so much as am presenting another option. Since the electron antineutrino has been overlooked in your hypothesis, there could be a more accurate way for this to unfold. The half integer spin would be a problem, as would the source of the antineutrino. It the fusion of two protons is to be symmetrical with the photofission of deuterium, then the neutrino should be included or accounted for otherwise. For radiation energy - I see the dividing line about what “would have been noticed” in the past 24 years of study as being in the range of 10 keV. Higher would have shown up, especially with glass electrolysis cells – lower than 10 keV could have been overlooked. From: Bob Higgins Jones, I think you did not understand or agree with what I said previously in bullet 4). On Wed, Jul 23, 2014 at 11:28 AM, Jones Beene <[email protected]> wrote: From: Bob Higgins Consider that the DDL state is regarded as being about 511 keV less than H in normal ground state. The mass energy difference between 2 ground state H atoms and a ground state D atom is 1.66 MeV… So, now the H#2 molecule may only be 1.66 - 2(.511) - (.1) = 538 keV different than the ground state D. Agreed. This 538 keV is still too large to go unnoticed without a step-down process but it does bring to mind the other possibility which itself is the downshifting mechanism itself – especially if the this DDL state is, in essence – dark matter. Mills and others believe this to be true. What I previously explained in 4) was that when the H#2 fuses, one electron ends up becoming part of a neutron (inverse beta) and the other electron is still in a fractional DDL orbital. When the nucleus gives off its residual 538 keV, it does so by giving it to the electron in that degenerate orbital. It will take 511 keV of the 538 keV to elevate the electron back to the ground state, so at that point, there is only 27 keV left in electron kinetic energy (in my previous post I made a stupid mental subtraction error and came out with 22 keV, but in this example, it is 27 keV). Since it only takes about 16 eV to ionize the atom, the electron continues on its way with essentially 27 keV of energy and the deuterium ion is left. I am not sure how and when the kinetic energy will be divided between the deuterium nucleus and the electron [Would the two only divide the 27 keV?]. Even still, this is much closer to the 3.5 keV x-ray in the dark matter. Bob H.
