Lasted edited by Andrew Munsey, updated on June 15, 2016 at 1:01 am.
Mosier-Boss and colleagues inserted an electrode composed of nickel or gold wire into a solution of palladium chloride mixed with deuterium or “heavy water” in a process called co-deposition. A single atom of deuterium contains one neutron and one proton in its nucleus.
Researchers passed electric current through the solution, causing a reaction within seconds. The scientists then used a special plastic, CR-39, to capture and track any high-energy particles that may have been emitted during reactions, including any neutrons emitted during the fusion of deuterium atoms.
At the end of the experiment, they examined the plastic with a microscope and discovered patterns of “triple tracks,” tiny-clusters of three adjacent pits that appear to split apart from a single point. The researchers say that the track marks were made by subatomic particles released when neutrons smashed into the plastic. Importantly, Mosier-Boss and colleagues believe that the neutrons originated in nuclear reactions, perhaps from the combining or fusing deuterium nuclei.
Before to understand how cold fusion occurs, there is need to understand how hot fusion occurs, as it is explained from the new nuclear model proposed in Quantum Ring Theory1.
Before to go on reading, we strongly recomend to read the Peswiki article: Article:Cold Fusion and Gamow's Paradox
The proton, according to QRT, is composed by:
1- A body ring
2- A principal field Sp(p) involving the body ring.
3- A secondary field Sn(p) involving the field Sp(e). Such field is responsible for Coumbic interactions. It has positive electric charge
The proton is seen in the Fig. 1.
The structure of the electron is similar:
1- A body ring
2- A principal field Sp(e)
3- A secondary field Sn(e). It has negative electric charge
When two nucleons are packed by hot fusion process, their secondary fields interact as shown in Fig. 2, where a deuteron interacts with a nucleus of oxigen.
CONDITIONS UNDER WHICH HOT FUSION OCCURS:
1- The two vector spins are always aligned (the two green arrows). This is the fundamental condition of the hot fusion process.
2- There is need a BIG energy able to win the repulsion between the two secondary fields (blue and red). Such energy must be able to make the overlap of the two fields (the blue field must be perforated). Within the Sun this big energy is caused by a big force due to gravitational pressure, and kinetic energy (the nuclei move fast due to the high temperature, and they collide with big kinetic energy).
Figure 2-A shows the 1H2 entering into the nucleus of oxigen, under the action of a force F, in the HOT FUSION process:
As we will show here, cold fusion occurs by the process shown in the Fig. 3:
CONDITIONS UNDER WHICH COLD FUSION OCCURS:
1- The two vector spins are situated along the same line (one arow green and the another pink).
This is the fundamental condition of the cold fusion process.
2- There is NO need a big energy, since the the repulsion between the two secondary fields (blue and red) is not so strong, because:
2.1- Although there is Coulombic repulsion between the two secondary fields blue and red in cold fusion process (Fig. 3), however its magnitude is very weaker than in the case of hot fusion process (Fig. 2), as we realize by comparing the two figures.
2.2- there is “hole” in the secondary field of oxigen (blue field). Therefore there is no need to perfurate the line fluxes of the oxigen secondary field (blue) in the cold fusion process, because the deuteron penetrates in the “hole”. So, there is no need that energy of perforation required in the hot fusion process.
So, we realize that cold fusion requires an energy very smaller than hot fusion. However, there is Coulombic repulsion between two deuterons, and so at the first glance it seems that a cold fusion D-D cannot, because there is not temperature or pressure able to promove their fusion.
Figure 3-A shows the 1H2 entering through the "hole" in the secondary field of the nucleus of oxigen, in the COLD FUSION process:
A nucleus Pd is composed by seven hexagonal floors, as shown in the Fig. 4 .
The distance “d” between two consecutives floors has contraction-expansion, a phenomenon named Accordion-Effect in Quantum Ring Theory.
In nuclei with complete hexagonal floors, like Ni, Pt, Au, Pd, when they belong to a sample, there is a tendency of such contraction-expansion to enter in resonance, when the vector spin of the nuclei is pointing out to the same direction.
Figure 4-A shows the sequence of contraction-expansion:
Typically, researchers working with these tabletop electrolytic experiments lower a palladium rod into a beaker of deuterated, or “heavy water," so named because it has a high concentration of deuterium.
When an electric current runs through the solution, the deuterium atoms pack into spaces in the palladium’s lattice like atomic framework, as shown in Fig. 5.
Over a period of days or weeks, the deuterium becomes packed, or "loaded," in densities of approximately one deuterium atom for each palladium atom.
Consider a nucleon of deuterium between two nuclei of Pd, as shown in Fig. 6.
The deuterion is submitted to a zig-zag motion because it has Coulombic repulsion with the two Pd nuclei, a the oscillation has the same direction of the accordion-effect of both the Pd nuclei.
Consider a second deuterion captured between two Pd nuclei, aligned with the first deuterion, as shown in the Fig. 7.
Let’s call them deuterium No. 1 and No. 2, as shown in the figure.
The red line indicates the trajectory of an electron orbiting the two deuteriuns.
Fig. 8 analyzes the attractions and repulsions considering three positions of the electron (shown as a black ball).
When the electron is situated in the point indicated in Fig 8-A, we realize that it eliminates the repulsion between the deuterium No.1 and the nucleus Pd above it. As the deuterium No.2 has repulsion with the Pd nucleus bellow it, the two deuteriuns go up.
Pay attention that it is variable the forces of repulsion caused by the Pd nuclei (shown in Fig. 8 with the green arrows), since they change their intensity because of the accordion-effect (when the Pd nucleus has dilation, the distance of its protons to the deuterium decreases, increasing the Coulombic repulsion, and when Pd has contraction the distance of protons and the deuterium increases, decreasing the Coulombic repulsion). So, the accordion-effect has a big influence in the resonance phenomenon.
In Fig. 8-B, the electron is between them. As the two deuteriuns are submitted to the forces of repulsion with the two Pd nuclei, and because the force of repulsion between the two deuteriuns decreases since the electron is between them, then we realize that there is a force trying to make a fusion between the two deuteriuns.
In Fig. 8-C again there is a force of repulsion between them.
Therefore we realize that the two deuteriuns are submitted to an oscillatory motion that sometimes tries to expel one each other, and sometimes it tries to make a fusion between them.
Note that the alignment of the oscillatory motion of the two deuteriuns along the same direction of the expansion-contraction of the accordion-effect is caused the magnetic moment of the nuclei.
So, it is very important to avoid deviations, as for instance by external magnetic fields. That’s why the magnetism of the Sun can influence in the problem of replicability, because it can deviate the deuteriuns from their oscillation along the contraction-expansion of the accordion-effect, and the resonance does cannot occur.
We realize that the two deuteriuns are submitted to a motion suitable to get resonance due to two things:
1- The accordion-effect of the two nuclei Pd
2- The velocity of the electron. Because depending of the velocity of the electron, the three positions indicated in Fig. 8 can resonate with the accordion-effect of the two Pd.
But theres a third cause of resonance, and it is very important.
Indeed, the electron moves with zitterbewegung. And the radius of the zitterbwegung can change with the velocity. Besides, when the electron passes between the two deuteriuns, the radius of its zbw tends to suffer a contraction. So, the zbw works like a spring, and a suitable temperature of the sample can help the resonence of the oscillatory motion of the two deuteriuns, in order that with the help of the resonance they can finally can one penetrate each other, and their fusion occurs.
When the otimum point of resoance occurs for two deuteriuns, they have fusion. A tritium is formed, with the emission of a proton with energy about 2,5MeV.
Note that the field Sn(Pp) of the nucleus Pd is the region where the electrons move in the electrosphere of the atom Pd. Therefore the proton will be attracted by the big negative field formed by the 46 electrons of the Pd electrosphere, in order that it gets acceleration, and emits X-rays during its motion with accelerated velocity. The proton’s trajectory is shown by the red arrow in the Fig. 9. The proton is shown in green, and the electrons are red.
Along its trajectory in the electrosphere of Pd atom, the direction of its motion is deviated several times by the proximity of some electron that passes near to it.
In such accelerated trajectory the proton can increase its energy from 2,5MeV to about 6MeV.
Finally the proton captures an electron, and they form a neutron. The electron loses its zitterbewegung, and the zbw energy is transmitted to the neutron, which now has a total energy of about 6 + 3,5 = 9,5 MeV.
The energy of zitterbewegung depends on the level where the electron is moving. In the deepest levels the zbw energy is greater.
The direction of the neutron’s motion depends upon the direction of the electron during the collision proton-electron (and also the position of the electron in its helical trajectory in the electrosphere of the Pd atom).
As the neutron has not charge, it moves by a rectilinear trajectory, shown by a black arrow in the Fig. 9.
If the neutron gets a direction toward the cathode Au, it will hit the sample CR-39. But it can also to take another direction, and then it does not collides against the CR-39.
The process continues. After the emission of the proton, a new deuterium is captured into the Pd lattice, and a partnership deuterium-tritium is formed. They get resonance by the same process, and a nucleus 2He4 is formed, with emission of one neutron with about 2,5MeV.
Depending on the conditions of resonance, the process is over with the fusion of 2He4, explained in the former item.
But if there are good conditions of resonance, two nucleons 2He4 can be formed together, as shown in Fig. 10.
The process of resonance is similar to that which occurred between two deuteriuns, but now the energy of oscillation is grater, because each 2He4 has two protons, instead of one as there was in the case of each deuterium.
2He4 particles leave the nucleus 92U238 with 4,2MeV, and so it can be expected that they can enter in a nucleus like Pd with smaller energy than 4,2MeV, by storing energy by resonance.
An email was sent to Pamela Mosier-Boss in 11 April 2009, suggesting to use an oscillator in her experiment.
The email is ahead.
From: Wladimir Guglinski (email@example.com)
Sent: Saturday, April 11, 2009 3:46:25 PM
Cc: firstname.lastname@example.org David Hestenes (email@example.com) EDEL PONS (firstname.lastname@example.org)
My theory can be tested by your experiment.
My idea is to use an oscillator capable to increase the oscillatory motion of the molecules D-D within the Pd lattice, by stimulating the resonance D-D.
If you succeed to stimulate the resonance D-D , we have to expect a growth in the rate of fusion D-D and also in the rate of neutrons emission by unity of time.
The oscillator I suggest is the following:
1- A glass buble is fulfilled by heavy hydrogen (D-D molecules).
The buble must be placed close to the Pd lattice deposited in the cathode.
2- Two electrodes are connected inside the buble.
3- A high voltage is applied to the electrodes, producing an electric discharge that crosses the gas of molecules D-D.
4- The molecules D-D into the buble are excited, and they emit photons in a frequency which is a sub-multiple of the frequency oscillation of the molecules D-D that fulfill the Pd lattice.
5- The molecules D-D within the Pd lattice get resonance with the frequency of emission by the D-D molecules into the buble, and the oscillation of D-D within Pd is stimulated to increase its amplitude (see the figure bellow).
6- I suppose such stimulation of resonance may increase the velocity of D-D fusion within the Pd lattice.
A SECOND ALTERNATIVE:
You can use a laser that hits the molecules D-D within the glass buble, instead of using an electrical discharge.
A THIRD ALTERNATIVE:
The best would be to build a laser which emission is produced by D-D molecules. In such case there is no need to have a glass buble, because the laser would be applied directly to the region of Pd lattice.
Perhaps you have to try the three alternatives.
It’s my opinion you should have to try it.
After all, we are in front to a new Physics, and we have to try any new idea if it makes sense.
Good luck in your attempt, if you decide to do it.
Subject: RE: absence of gamma-rays in your experiment, and neutron’s background
Date: Mon, 13 Apr 2009 10:29:47 -0700
CC: email@example.com firstname.lastname@example.org email@example.com
Like many, we have very few funds and resources. But we will consider your suggestions and see what we can do as time and money permits.
Script on Quantum Ring Theory:
1- W. Guglinski, Quantum Ring Theory- Foundations for Cold Fusion, Bäuu Press, 2006