On May 31, 2013, at 12:11 PM, Bob Higgins wrote:
Dear Dr. Storms,
Yours is a fascinating theory, but I don't understand the mechanism
you propose of "slowly reducing the Coulomb barrier" by photon
emissions from the nucleus.
We start with two protons each having a charge of 1. We end with
single deuteron having a charge of 1. Consequently, a charge of 1 has
to disappear, which it does by reacting with the intervening
electron. In addition, we start with a mass excess, which has to be
converted to energy. I'm proposing that the unit charge is retained.
Because the two protons have to combine into a single nucleus at the
end of this process, the distance HAS to be reduced to zero at some
time during the process. I prefer to believe this reduction is gradual
as mass-energy is lost. For this to happen, I propose the effective
barrier is reduced as mass is converted to energy of photons. The
protons gradually get closer only because the barrier is reduced, not
because the unit charge is changed.
I agree, the details can get complex. My first goal is to get the
basic process understood. Then we can discuss details. I'm sure the
details will cause some changes in the description, but the basic
approach I think is important and needs to be understood. I'm glad
you find it fascinating, Bob. That is the first step.
Ed Storms
The Coulomb barrier, as I understand it, is the proton-proton
electric field repulsion between the hydron elements of the Hydroton
molecule. Each proton has a unit quantized positive charge, so I
presume the coulomb barrier reduction is not coming from reduction
in the charge of the proton (you are not proposing fractionating the
unit charge, or are you?). I gather the electron orbitals in the
Hydroton are screening the charge on the neighboring protons. If
the Coulomb barrier is being reduced, I can imagine screening of the
charge of the protons by a change in electron orbitals. This is now
sounding a little like Mills-ian fractional Rydberg change in the
orbital, allowing the electron wave function to shrink closer to the
proton which provides a screening until protons are closer
together. Perhaps the electron orbital becomes squashed like a disk
where it orbits very closely along the hydroton axis around the
proton and extends way out into the walls of the NAE crack.
However, if this were the case, then the photons corresponding to
the Coulomb barrier reduction would be coming from orbital
transitions of the electron and not from the nucleus.
Are you instead suggesting some kind of proton valence quark
oscillation that would make the proton appear like a neutron for
some fraction of the time? (A naive guess on my part I am sure.)
Can you provide additional insight into your proposition?
Regards,
Bob Higgins
On Fri, May 31, 2013 at 11:37 AM, Edmund Storms
<stor...@ix.netcom.com> wrote:
As this mass-energy is reduced, the Coulomb barrier is lowered
further, permitting the two nuclei to get closer at each cycle. Once
the nuclei fuse, the Hydroton ceases to exist and instead nuclei of
D are present if the original nuclei in the Hydroton were H, the
final nuclei is He if D made the Hydroton, and the final nuclei is
tritium if H+D were in the Hydroton. The He diffuses away while the
tritium and D can enter other Hydrotons that continuously form. This
is a contineous process limited ONLY by how fast the hydrogen
isotopes can get into the gap.
No, the Coulomb barrier is slowly reduced in height as mass-energy
is lost, thereby allowing the nuclei to get closer each time the
cycle repeats. Finally, the Coulomb barrier disappears and the two
nuclei fuse, but very little excess mass-energy is present when
this happens. Consequently, when the electron is absorbed, the
resulting neutrino has very little energy to carry away.
