Keep in mind as well that Holmlid adduces not only muons, but kaons and
pions as well. Once we introduce (negative) kaons, we have the following
decays to deal with:
[image: Inline image 1]
The neutral pion assures us that there will either be penetrating gammas or
positrons, which lead to 511
I believe that when the muon decays, if it is a negative muon, it decays
into an electron and a pair of neutrinos. If it is a positive muon, it
decays into a positron and 2 neutrinos. Before it decays, if it enters the
electronic structure of an atom (likely in condensed matter), then it
quickly
Hi Fran,
I am unable to imagine how something special would happen in that case. A
muon in slow motion may have a greater chance of interaction if its energy
is near the ionization energy of the atoms upon which it is incident - but
this is only a small energy - less than 10eV. At higher
Physicists Observe Rydberg Molecule for The First Time
http://flip.it/s9Qrqn
What is the mode of "decay" of free muons and, separately, in condensed matter?
They seem not to produce any high energy EM nor radioactive products. If they
did, I would assume this would have been reported unless it was intended to
remain a secret.
I consider based on reported muon models
Axil's post is one interpretation of QM, other could be that the QM fields
represents real fields e.g. no particles in space. This means that you can
view QM as billiard with fields in stead of balls and things get to be much
less mystic. Also Mills is starting to get real evidences of over unity
We are talking Quantum Mechanics here, not billards. In QM,
superposition means that the muon can be in many places at once while
it is in the entangled state. Distance does not matter. Where the muon
ends up is based on decoherence of what has entangled the muon with
the LENR reaction. It is all
Bob, what if the “muon” doesn’t have to achieve light speed but rather becomes
so “suppressed” think traveling thru a tiny Casimir cavity that the muons
actual speed inside the cavity where vacuum wavelengths are dilate by
suppression appears to achieve negative light speed relative to
http://egooutpeters.blogspot.ro/2016/11/nov-14-2016-lenr-info.html
peter
--
Dr. Peter Gluck
Cluj, Romania
http://egooutpeters.blogspot.com
The idea that the muons are interacting in solid matter with the electrons not
the nuclei of atoms is very compelling to me. Indeed this may well explain two
mysteries of my cold fusion muon/mischegunons, that is that very few are
escaping the experiment cells. That what I have detected is the
Bob,
You are conflating two or more different Holmlid papers…
He is unambiguous. The 10MeV particles are clearly stated to be “mainly protons
from the fusion process and deuterons ejected by proton collisions” (see the
abstract from the paper).
The muon observations are from other
From: Bob Higgins
In this discussion, Jones presumes muons to be traveling at light speed…
That is an oversight. It should read “a large fraction of light speed”…
.
In this discussion, Jones presumes muons to be traveling at light speed:
The muon is an unstable fermion with a lifetime of 2.2 microseconds, which
is an eternity compared to most beta decays. Ignoring time dilation, this
would mean that muons, travelling at light speed, would be dispersing and
Hi all
EDITED TO CORRECT AN ARRITHMETIC ERROR!
~600 m not 6000m
By the way the density of the incidents has to be distributed across a
sphere that is approximately 1,440,000 π (pi) meters squared.
Then you have to plug in the distribution curve to get cubed meters for
area.
The numbers are
Hi all
By the way the density of the incidents has to be distributed across a
sphere that is approximately 144,000,000 π (pi) meters squared.
Then you have to plug in the distribution curve to get cubed meters for
area.
The numbers are very big
Hence why I think the density will be very small.
Hi all
With that size of sphere, 6000m radius, I am guessing, from experience the
density of interactions will be only a little above natural background. You
need to know the surface area of the sphere. Then the distribution curve
for the straight line from the source; then calculate peak and the
16 matches
Mail list logo
