Jones,
I also do think that there could be more optimal wave length. (they
probably know it.. how to get the right negative muons).
The calculation shows that the imputing energy (wave - topology) of
polarized photons is in fact homomorphous to the so called 1x1 X (1x1 =
back side) 1FC orbit or simply the symmetric coupling of two identical
waves or a resonance of it. (resonances of a wave are topological
allowed integer fractions)
The proton 3D/4D perturbative mass is indeed = 11.4MeV (finally 11 MeV
for scattering) what is also the sum of the measured quark masses for
the 3 neutral waves. The perturbation is given by the coupling (don't
forget the 2-->3 weight!) with the 272kev potential (1x1) generating
waves and to smaller extent with the relativistic ( 4 rotation) core
mass. (what finally gives 11Mev)
As said: All magnetic waves with the same topology can potentially
couple. (What SM believes to be virtual particles...). Such a coupling
is immediate and not restricted by the speed of light. (Same as spooky
action on a distance). If you hit the soft spot then - bang.
With a laser you just add energy to a (1x1) orbit - the visible side of
the orbit. Due to overloading of the field - stored energy - cannot
relax, as the mean relaxation time is much longer than the delayed
arrival of the next photon. It's like incrementing a counter.
But there is one trick. The 8H* --> 2 4-He (or 8-Be) conversion to Kaons
is exothermic ( not for 4Deuterium*) . Thus the added energy is not
delivering the Kaon split energy. The added energy that synchronizes
with the 2 potential waves and indirectly with its multiple the (2x2)
core wave *increases the magnetic moment* what induces a stronger
coupling of the 8H*. (what also Mills claimed..)
In H* the proton 3D/4D perturbative mass is 1FC paired. The 8H* --> 2
4-He process exactly releases 4 3D/4D perturbative masses + 4 (4-He)
internal deuterium bond energies.
So we can draw the following picture: When the resonant 8H* coupling
energy hits the 1/7 proton (a ninth proton!) wave excess energy then a
proton can resonantly take it over and splits as it converts from 3x3 9
waves to 8 - 2x2 .. 2x2 + 1/7. or finally 2x2 + 1/14 .. 2x2 + 1/14.
Whether the Kaon has a ("long-time") stable wave structure or not has to
be modeled and proven by better experiments.
The Holmlid process is the highlight of this centuries physics as it
shows a new path to physics and how to model it. It also enables a
terribly cheap table top production of Kaons,Pions, Muons. And in fact
the fraction of -u depends on the setup! An of course the production of
Helium has been confirmed albeit it is obvious from the SO(4) physics model.
J.W.
Am 05.02.20 um 18:16 schrieb Jones Beene:
Jürg
This is very interesting assuming one can use this information to
engineer proton disintegration with minimal input energy.
To that end, it would seem necessary to know the resonance wavelengths
in question (or frequency of the 1/7th and 1/9th waves). From that
information, one could presumably try to maximally disrupt that
resonance, possibly with a beat wave.
Would this be the basic 53 MeV resonance you mention or is there a
lower value which works?
In the standard model, the scattering cross-section of the proton is
around 1.5 fm (or 11 MeV) IIRC so there is a big gap there with
available lasers.
Presumably Holmlid is doing this kind of disintegration with a laser.
Holmlid may have stumbled onto an effective wavelength which is not
optimum. Who knows? Perhaps his laser somehow stimulates a much
shorter wavelength.
Jones
----------------
Jürg Wyttenbach wrote:
> The allowed torus resonances are 7 and 9 waves. The proton base
state has 9 waves.... The Holmlid proton split seen from the proton
is: One out of 9 proton waves starts the an orthognal 1/7 resonances
what leaves behind a (2x2)x(2x2) wave structure without the biding
glue of the 3D/4D waves. This wave pack (4x4) is repulsive as seen in
8-Be. The basic energy 53MeV for the split is delivered from the
resonant 8H* --> 2 4-He (or 8-Be) conversion.
--
Jürg Wyttenbach
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