On Jul 24, 2007, at 10:12 AM, Jones Beene wrote:
[side note]...wonder if one might present the premise that the
Cooper pair is the most basic form of degenerate electron ?
Yes, in the form of the 2e- boson, a very weekly bound version. Deja
vous! I think we had this discussion once Jones. Ah yes here it is:
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----- Original Message ----- From: "Horace Heffner"
Such a process may well provide a mechanism for electron boson
formation and for energy focusing phenomena in the anode interphase.
That is an interesting paper you have up at:
http://www.mtaonline.net/~hheffner/PairLNR.pdf
Very provocative... and having been thinking lately about the
features, and real identity (and past mis-characterization) of
Dirac's monopole, the possibility might exist for some overlap with
the characteristics of the e-boson ...?
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On Jan 8, 2006, at 12:12 PM, Jones Beene wrote:
Horace
The the 2e- boson, due to full superposition and opposed electron
spin, has no magnetic field. A monopole with spin would have an
electric dipole field. If non-zero spin monopoles exist, of the
Dirac kind or otherwise, then they might be able to form a
monopole boson having no electric dipole field.
Do you see any difference with the 2e-boson and the Cooper pair?
Yes. The Cooper pair electron waveforms need not be co-centered.
They merely exhibit long range coherence of their momentums. The 2e-
boson concept has the added condition that the waveforms are fully
overlapped, but with spins opposed. They are energetically slightly
bound by their magnetic fields when they are slightly perturbed.
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On Jan 8, 2006, at 1:22 PM, Jones Beene wrote:
Hmm. I'm having a hard time getting the mental picture, unless the
spins are orthogonal to each other (two axis spin)... but you are
saying that is not the case and the waveforms can overlap except
for spin, which is opposed, so that it cancels the magnetic moment
until the pair is perturbed, is that correct ?
Yes. Further, provided the magnetic binding energy is not exceeded
by the coulomb force, the force between the electrons is attracting.
What happens to the angular momentum? This seems to deny
conservation of angular momentum.
No problem there. 1 - 1 = 0 All balances before and after pair
formation.
Is the net charge as felt by other particles doubled?
Yes. However, in a superconductor, the waveform is very large,
spread over many nuclei. The net charge in a superconductor is
zero. When tunneling across a Josephson Junction, electrons
(empirically) tunnel in pairs roughly as often as not. This to me
indicates the biding energy is effective in creating a single event.
It seems to me logical the tunneling of a pair to separate locations
would be separate events. The target of the tunneling then is
logically a single location. The constituants of such a combined
tunneling thus also seem more likely to be bound as a single
waveform. Given that the waveforms are fully co-centered, then there
can be no Coulomb repulsion. Further, the magnetic field of opposed
spin particles can maintain that lack of Coulomb repulsion. The
center of charge of such a boson is a prospective locus for further
tunneling of (the center of charge of) more such Bosons. Thus you
have a mechanism for creating Ken Shoulder's condensed charge. The
center of charge of the 2e- boson is similarly an ideal locus for the
simultaneous tunneling of a pair of short-range cohered nuclei, like
possibly the two hydrogen nuclei of the same water molecule. The
powerful electrostatic field of the interphase brings the hydrogen
"ears" of the water molecule close together. When such a water
molecule pops, the two hydrogen nuclei may be especially receptive to
a new 2e- boson home. It just happens electrons from the oxygen
directly adjacent to and upstream from the "ears" is very willing at
that instant to provide such a home.
Horace Heffner
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See the thread "What's the story with e-bosons?"
Horace Heffner
http://www.mtaonline.net/~hheffner/
