The goal is to try and find a way in which charge screening can fission a
nucleus. What factors cause the nucleus to fall apart into many smaller
pieces?



In order to explain how charge screening can make a nucleus unstable, we
need to answer how the proton can transmute into a neutron. Neutrons are
heavier than protons. Where does the additional mass come from when a
proton is transmuted into a neutron?



To answer this question, let’s define some terms. Two terms are used in
referring to a quark's mass: current quark mass refers to the mass of a
quark by itself, while constituent quark mass refers to the current quark
mass plus the mass of the gluon particle field surrounding the quark.



These two masses typically have very different values. Most of a hadron's
mass comes from the gluons that bind the constituent quarks together,
rather than from the quarks themselves. While gluons are inherently
massless, they are both numerous and they possess energy—more specifically,
quantum chromodynamics binding energy (QCBE)—and it is this energy that
contributes so greatly to the overall mass of the hadron.





For example, a proton has a mass of approximately 938 MeV/c2, of which the
rest mass of its three valence quarks only contributes about 11 MeV/c2;
most of the remainder can be attributed to the gluons' QCBE.





When the nucleus is surrounded in a sea of numerous screening electrons, a
few of these electrons will find their way into the nucleus.



The transformation of a proton into a neutron inside of a nucleus is
possible through capturing one of these electrons by a proton.



Hey, this is starting to sound like the L&W theory. But the electrons don’t
need to form outside the nucleus. They get inside to do their
mischief…which is the destabilization of the nucleus.





Cheers:  Axil


On Sun, Jul 1, 2012 at 9:11 PM, Axil Axil <janap...@gmail.com> wrote:

> Repulsive interactions between neutrons
>
>
> To set the stage, the standard model of particle physics says that
> neutrons attract each other.
>
>
> From an idea from an outlier (our kind of people) in the field of physics
> Dr. O, Maneul as follows:
>
>
> “Neutrons and protons in the nucleus work like the north and south ends of
> magnets,” Manuel explains. “Neutrons repel neutrons, protons repel protons,
> but neutrons attract protons. Neutron repulsion is the force that energizes
> neutron stars. This empirical fact was discovered by five graduate students
> working with me to decipher the nuclear mass data for the 2,850 known
> nuclides in the spring of 2000.”
>
> http://www.omatumr.com/abstracts2004/nuclearclustering.pdf
>
> Nuclear clustering and interactions between nucleons
>
> Other papers by Dr. Maneul as follows:
>
> http://www.youtube.com/watch?v=sXNyLYSiPO0
>
> http://arxiv.org/pdf/1102.1499v1
>
> http://www.omatumr.com/abstracts2003/jfe-neutronrep.pdf
>
> My theory about how electrostatic screening of the nucleus can produce
> energy through nuclear fission and neutron breakup.
>
> To start out with, we cannot keep loading neutrons into a nucleus without
> limit because they will eventually be repelled by a mismatch with the
> resident collection of bound protons. So, more or less, protons and
> neutrons must be paired inside a nucleus. Any mismatched nucleon is
> expelled from the nucleus.
>
> In Cold Fusion, when all the positive charges of the protons are screened
> from the nucleus by a large collection of electrons, Could it be that the
> protons become neutrons by catalyzing a color change though gage boson
> mediation. The strong force in the nucleus becomes now repulsive and the
> nucleus and all the nucleons in the nucleus become unstable. This screened
> nucleus then begins to fall apart because of the short-range repulsive
> neutron–neutron interactions mediated by the strong force.  The neutrons
> now mostly converted to neutrons in that nucleus begin to disintegrate into
> hydrogen atoms.
>
> From  Dr. Maneul: The disintegration of a neutron can generate up to 3% of
> its rest mass into energy.
>
>
> This expositon say it far better than I can. The case for neutron
> repulsion as follows:
>
> http://www.applet-magic.com/neutronrepulsion.htm
>
>
> Cheers:  Axil
>
>
> On Sat, Jun 16, 2012 at 11:05 PM, David Roberson <dlrober...@aol.com>wrote:
>
>> Earlier I made a posting about the addition of a proton or neutron to a
>> stable isotope and observed that if one of these new compositions is stable
>> then the other one must not be.  This observation holds throughout the
>> entire list of elements on the chart that I have been referencing.  Now
>> I have a hypothesis as to why this is true.
>> First of all, it is important to note that the above additions result in
>> a new element or isotope that has one additional nucleon.  After
>> completing a great deal of research on the subject I see that a group of
>> elements that share the same number of nucleons have an interesting
>> behavior.  They exhibit energy levels like an electron cloud around a
>> single nucleus.  A minimum energy level (ground level) is always present
>> and adjacent elements are always at a higher level.   It takes one beta
>> plus or minus decay to get between these adjacent levels and it appears
>> that this will be energetically favored and always occur at some future
>> time.  The time frame for this decay might be quite extensive, but it
>> will be measurable in the form of radioactivity.  I think of this
>> process as a lot like the decay of electrons from higher energy levels
>> which eventually get to the ground state.
>> I constructed an equation that can be used to find the expected number of
>> nucleons as a function of the number of protons within a nucleus.  I
>> restricted the range of protons so that it eliminates the very few proton
>> case and also stops at a proton count of 40 so that I can concentrate the
>> research to the region to which I am interested and to improve the curve
>> fit substantially.
>> I then transformed the above expected nucleon number verses proton count
>> into the reverse relationship.  In this manner I can enter the nucleon
>> count and calculate the proton and neutron numbers that ideally support it.
>> As an example, if I enter a nucleon count of 40 I arrive at an expected
>> proton count of 19 and an associated neutron count of 21.  The element
>> that this chooses is potassium 40.  Now this should be the location of
>> the minimum energy level or ground state for 40 nucleons.  The
>> interesting thing I observed is that this element is unstable with a very
>> long half life.  My explanation is that the number of protons and the
>> number of neutrons are both odd so they cannot pair up.  The resulting
>> mismatch reduces the binding energy enough that it actually falls below the
>> adjacent elements of Calcium 40 and Argon 40 which each have an even count
>> of both types of nucleons.  Also, it is apparent that the fall off rate
>> of binding energy verses error in nucleon ideal distribution is such that
>> the next element on each side of the two above has less binding energy than
>> these ideal ones.  For this reason I propose that a beta type decay
>> process will eventually yield one of the two stable levels one step at a
>> time as each decay takes place.
>> It should be noted that this odd proton, odd neutron count situation is
>> virtually always unstable.  And likewise the even proton, even neutron
>> case is similarly always stable when it occurs at the ideal count position.
>> When just one of these counts is even the elements tend to be stable, but
>> less so.
>> I have not reviewed the cases where other types of decays are present and
>> that might yield fertile ground for future research.
>> The conclusion I draw from my search is that there will not be a case
>> where the addition of a proton or a neutron from a currently stable element
>> will both result in a stable new isotope or element.  One or the other
>> of these processes has less binding energy and a beta plus or beta minus
>> decay will point to it.  There are conditions where neither new product
>> is stable, but these are fairly rare.
>> I plan to analyze the decay times of the energy steps in these nuclear
>> configurations and compare them to the time frames for electron energy
>> level steps.   The levels of energy events within the nucleus are
>> enormously larger than chemical ones but the decay times are likewise much
>> longer.  These characteristics seem to be contradictory.
>> Dave
>>
>>
>>
>>
>

Reply via email to