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Cold fusion mystery finally deciphered

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After the Lugano Report Observation of abundant heat production from a reactor device and of isotopic changes in the fuel , published in October 2014 on the results of the Andrea Rossi cold fusion reactor, the mystery of the cold fusion reactions was finally deciphered, by W. Guglinski.

The Lugano Report can be found in the link:

http://www.sifferkoll.se/sifferkoll/wp-content/uploads/2014/10/LuganoReportSubmit.pdf

The mechanisms for the cold fusion were explained in the Andrea Rossi’s Journal of Nuclear Physics, where some questions were responded for the readers of the JoNP.

The discussion started in this link:

http://www.journal-of-nuclear-physics.com/?p=864&cpage=6#comments

How cold fusion may contribute for the solar nucleosynthesis

All the current theories of Modern Physics had been developed from the concept of empty space. So, the concept of field considered in the QFT–Quantum Field Theory does not take in consideration any structure for the space.

That’s why in the Standard Nuclear Physics the electric field of the particles as the proton, the electron, and also the electric fied of the nuclei, is considered as a spherical field involving the particles, or the nucleus.

This electric field of the nuclei considered in the Standard Nuclear Physics is a homogeneous sphere (it means that in any point of the field the value of the electric vector is always the same). Therefore, when a particle as the proton is forced to enter within a nucleus because it is submitted to very high pressure and temperature, the energy necessary to win the Coulomb repulsion is always the same (because no matter where is the point of the electric field of the nucleus where the proton enters, the energy necessary for the proton to succeed to enter is the same in any point of the electric field of the nucleus).

Such concept of field adopted in the QFT introduced several puzzles in the Standard Nuclear Physics.

For instance, when an alpha particle (2He4) exits the nucleus 92U238, it exits with an energy lower than the energy necessary to put an alpha particle within the nucleus 92U. Such paradox was solved by Gamow. However his solution is not acceptable, because he introduced another paradox in his solution. Besides, if the 2He4 should exit the 92U as proposed by Gamow, because the electric field of the 92U is spherical the 2He4 would have to exit the 92U with a tangential line (because of the rotation of the 92U). But the experiments show that the 2He4 exits the 92U with a radial line.

Other puzzle is the emission of solar neutrinos by the Sun, as we see from the paper published by the journal Nature in 1984:

Solar neutrinos and other problems and their relation to energy production in the Sun:

http://www.nature.com/nature/journal/v312/n5991/abs/312254a0.html

“Models of the solar interior, based on the usual physical assumptions, predict a neutrino flux several times greater than that observed in the Davis 37Cl experiment. If, as is widely accepted, this discrepancy represents a ‘flaw’ in the standard solar model one would expect this flaw to manifest itself in other ways also. Here we point out some less well known discrepancies between theoretical predictions of the standard solar model and relevant observations. ”.

The problem is not solved yet, as we realize from the last updated on 23 September 2013 version of the paper The 3He(?,?)7Be reaction for big bang and stellar nucleosynthesis:

http://www.york.ac.uk/physics/research/nuclear/nuclear-astrophysics/big-bang/

”… looking for physics beyond standard model of particle physics. Naturally, the reaction attracted early attention of experimentalists and theoreticians alike in the 1950’s. Surprisingly, even today much work needs to be done via nuclear physics experiments to understand this reaction and provide information to the colleagues working on big bang nucleosynthesis, standard solar model and standard model of particle physics”.

Perhaps the puzzle of the rate of solar neutrinos cannot be solved via the Standard Model because they are not considering the cold fusion in the Sun. The emission of neutrinos from cold fusion reactions occurs in a rate very lower than that occurring in hot fusion.

Probably some steps in the nucleosynthesis of some elements in the Sun occurs via cold fusion. And therefore it is impossible to conciliate any theory developed from the Standard Model with the experimental astronomical observations.

An experiment published in 2011 proved that space is no empty

The experiment was published in the journal Nature. Light was produced by the space. And therefore the space cannot be empty. It must have a structure, so that to be able to emit light.

A vacuum can yield flashes of light:

http://www.nature.com/news/a-vacuum-can-yield-flashes-of-light-1.12430

Therefore the concept of electric field adopted in the Quantum Field Theory must be wrong.

A new concept of electric field, based on the concept of a space having a structure is proposed in the Quantum Ring Theory.

And here a crucial point emerges: there is no way to propose a spherical shape of electric field by considering the space with structure.

The concept of field adopted in Quantum Field Theory cannot be correct

Other fundamental puzzle impossible to be solved by considering the concept of field adopted in QFT is concerning the null magnetic moment of even-even nuclei with equal quantity of protons and neutrons, as 2He4, 4Be8, 6C12, 8O16, 10Ne20, etc. Due to the monopolar nature of the electric charge. For instance, the 2He4 has two protons. As the nucleus has rotation, the rotation of the two protons has to induce a magnetic moment. So, by considering the model of field adopted in QFT, it is impossible to explain the null magnetic moment of the 2He4, and all the other even-even nuclei with equal quantity of protons and neutrons.

I had challenged several nuclear theorists for coming to Rossi-Focardi blog Journal of Nuclear Physics-(JoNP) so that to explain how such puzzle could be solved according to the Standard Model. No one nuclear theorist did come.

The reason why the model of field adopted in Quantum Field Theory cannot explain the null magnetic moment of the even-even nuclei with equal quantity of protons and neutrons is because in QFT it is adopted the mono-field model. In my paper Aether Structure for unification between gravity and electromagnetism it is shown that the null magnetic moment of those nuclei can also be explained via a double-field model (an outer electric field concentric with an inner central field composed by gravitons), adopted in Quantum Ring Theory.

The paper Aether Structure for unification between gravity and electromagnetism was published in the Rossi's Journal of Nuclear Physics, and it can be found in this link:

http://www.journal-of-nuclear-physics.com/files/Aether%20Structure%20for%20unification%20between%20gravity%20and%20electromagnetism.pdf

The Fig. 1 ahead shows the two concentric fields of a proton, as proposed in the paper Aether Structure for unification between gravity and electromagnetism.

FIG. 1

Image:FIGURE 1- 3 fields of the proton.png

As the radius of the electric field has the magnitude of the Bohr’s radius 10^-11m, and the radius of the nucleus is 10^-15m, of course the Fig. 1 does not show the real proportion between the fields. The Fig. 2 show a better proportionality (but of course not real yet):

FIG. 2

Image:FIGURE 2- 3 fields in real proportionality.png

As seen in the Figure 2, there are two “holes” in the electric fields of the particles, and also in the electric fields of the nuclei.

Under suitable condictions of low pressure and temperature, a nucleon as a proton or a deuteron can enter within a nucleus through that hole by having lower energy than that necessary if the nucleon is forced to enter via any other point of the electric field of the nucleus.

By considering that nucleons also may exit a nucleus via the hole in the electric field of the nuclei, we eliminate two paradoxes of the Standard Nuclear Physics:

1) The unacceptable paradox introduced by Gamow, proposed for explaining how a 2He4 can exit the 92U with energy lower than the necessary to cross the Coulomb barrier of the electric field of the 92U

2) Why the 2He4 exits the 92U by a radial trajectory as detected by experiments (impossible to explain from the Gamow theory based on the Standard Model).

How a nucleon may enter within a nucleus

Cold fusion may occur by two ways:

1) Via resonance within vessels with conditions of low pressure and temperature, as occurs in the Rossi’s E-Cat.

2) Via kinetic energy in vessels with conditions of very high pressure and temperature, as the Sun. Let us see how it may occur:

a) In the Sun, more than 99,999% of the fusions occur via high nuclear reactions.

b) Cold fusion occurs in less than 0,001% of the nuclear fusions

c) It is very hard to occur cold fusion in the Sun, because the nucleons (for instance a proton) have very high kinetic energy in the star. So, when a proton enters within a nucleus via the “hole” in the electric field of the nucleus, the fusion does not occur (the nucleus cannot capture the proton) because due to the very high kinetic energy the proton simply trespass the nucleus, exiting the nucleus in the other “hole” opposite to the “hole” where the proton had entered.

d) But a cold fusion reaction may occur as follows:

d.1) Within the Sun all the nuclei are moving very fast, and every time changing the direction of their motion due to the collision with other nuclei.

d.2) But suppose that a nucleus (for instance 3Li7) in an exact instant is moving along the x-axis with speed “v”, with the “hole” of its field aligned toward the x-axis. And consider that a deuteron with speed “V” (moving in the same direction along the x-axis) in that exact instant collides against the 3Li7, because the speed V is faster than v. In that condition the relative kinetic energy of the deuteron regarding the 3Li7 is E= 0,5.m.(V – v)2, where “m” is the mass of the deuteron. Therefore the kinetic energy of the deuteron in some very rare conditions is suitable low so that, when the deuteron enters within the 3Li7, the deuteron is captured, and 3Li7 transmutes to 4Be9.

The difference between cold fusion and hot fusion

There are three basic differences between hot fusion and cold fusion:

1) Hot fusion occurs when a nucleon enters within a nucleus (only under extreme conditions of high pressure and temperature) when the nucleon succeeds to perforate the electric field of the nucleus.

2) Cold fusion occurs when the particle enters within a nucleus via the “hole” existing in the electric fields of the nuclei. It can occur either in low or in high conditions of temperature and pressure.

3) In order to enter within a nucleus via hot fusion, a particle needs to have a very big kinetic energy. So, when the particle enters within the nucleus, due to the very high kinetic energy of the nucleon the nucleus is excited, and this is the reason why gamma photons are emitted. Unlike, in the case of cold fusion, as the particle enters with low energy the nucleus is not excited, and gamma rays are not emitted. So, also the tax of neutrinos emission in cold fusion is lower than in the case of hot fusion.

Therefore, such property of cold fusion of emitting lower quantity of neutrinos can be response for the question why from the Standard Nuclear Physics there is no way to conciliate the hot fusion reactions in the Sun with the flux of neutrinos emitted by that star.

Conclusion

As we may realize, beyond the challenge of finding a theory capable to explain cold fusion by keeping the principles of the Standard Nuclear Physics there are many other challenges in Nuclear Physics impossible to be solved via the Standard Nuclear Physics.

Up to now the nuclear theorists refused to think about a New Theory based on new fundamental concepts missing in the Standard Nuclear Physics, because of two reasons:

1) Cold fusion is impossible to occur by considering the Standard Model

2) Therefore they were sure it would be possible to solve the unsolved questions by keeping the Standard Model.

But now cold fusion is a reality: Rossi’s Effect was confirmed by three universities of the Europe. And the nuclear theorists worlswide are beginning to accept this new reality, as by one of the top level nuclear phusicist of Russia, Dr Uzikov:

http://www.proatom.ru/modules.php?name=News&file=article&sid=5595

Therefore a reasonable person must realize that it makes no sense to continue trying to explain cold fusion via the old Standard Model. Because the problem is beyond the challenge of finding a theory for explaining cold fusion, actually now the cold fusion became the way for solving the unsolved questions.

Some nuclear theorists have the hope to explain cold fusion via the Standard Model. When we reply to them that cold fusion is impossible to occur from the principles of the Standard Model, they say that we don’t know in deep the Standard Model. However, we may reply to those nuclear theorists: actually you don’t know in deep the structures existing in the Nature. They are very different of that considered in the Standard Nuclear Physics, by beginning from the structure of the space, and as consequence the structure of the electric field of the nuclei, responsible for the difference between hot fusion and cold fusion.

On the nature of the force binding protons and neutrons within nuclei

• Wladimir Guglinski

November 19th, 2014 at 5:55 PM

Andrea Calaon wrote in November 19th, 2014 at 10:35 AM:

''“I am convinced that these are the reactions Andrea Rossi is investigating (remember his excitement at the article speaking about nucleus tunneling between Lithium7 and Nickel?). Note that, if the nuclear binding energy is not related to the nuclear force (as suggested by many and as in my theory), these reactions do not even involve the Weak interaction: They are purely electromagnetic.

Here are the reactions and the energies:

Ni58+Li7 ->Ni59+Li6 + 1.749 [MeV]

Ni59+Li7 ->Ni60+Li6 + 4.138 [MeV]

Ni60+Li7 ->Ni61+Li6 + 0.570 [MeV]

Ni61+Li7 ->Ni62+Li6 + 3.346 [MeV]” ''

------------------------------------------------

Dear Andrea Calaon

Actually we have to be astonished with the question: why did not the nuclear theorists realize 80 years ago that strong nuclear force cannot be responsible for the nuclear binding energy of the nuclei??? Because if the strong force was interaction which responsible for the attraction proton-neutron and neutron-neutron, then the dineutron would exist in the nature.

Two neutrons linked by the strong force cannot be separated by the isospin proposed by Heisenberg, because only a force of repulsion would be able to win the attraction by the strong force between two neutrons, and an abstract mathematical concept as the isospin cannot create a force of repulsion.

So, from a simple question of logic, the strong force must be discarded as the responsible for the attraction proton-neutron and neutron-neutron within the nuclei.

But the question is not so easy as it seems.

By considering the Coulomb repulsion in the distances of 1fm between protons and neutrons within the nuclei, the electromagnetic interaction is not able supply the necessary force for the agglutination of the stable nuclei. There is need an interaction 100 times stronger than that promoted by the electromagnetic interaction in a distance of 1fm, and this is the reason why the nuclear theorists had discarded 80 years ago the electromagnetism as the promoter of the nuclear binding energy, and they had adopted the strong nuclear force, which interaction is 100 times stronger than the electromagnetism in the distance of 1fm.

And now finally the E-Cat is showing what the logic was suggesting to us, when we had faced the obvious: as two neutrons do not form the dineutron, then the strong nuclear force cannot promote the agglutination of the nuclei.

And the consequence: the nuclear theorists will be obliged to accept this unavoidable fact.

Nevertheless a problem arises: as the nucleus is not bound via the strong nuclear force, but in the distances of 1fm there is need a force 100 times stronger than that promoted by the electromagnetism, how can the nuclei be bound via the electromagnetism?

Obviously an acceptable new nuclear model must be able to explain such paradox, and the nuclear theorist will discard the theories which do not solve the puzzle.

You said: “as suggested by many and as in my theory”.

However, how do you (and the many) explain it ?

In the nuclear model proposed in my Quantum Ring Theory the puzzle is solved as follows:

1) The nuclei are surrounded by an electric field.

See Figure 1:

Image:FIGURE 1- 3 fields of the proton.png

2) Suppose a proton fuses with a nucleus. The fusion occurs as explained ahead.

3) The proton must perforate the electric field of the nucleus, so that to be captured by the nucleus.

4) When the proton perforates the electric field of the nucleus and they have fusion, the electric field of the nucleus has no repulsion with the electric field of the proton, because the two electric fields fuse by forming one unique electric field, surrounding the nucleus and the proton. By this way, the proton is not submitted to that Coulomb repulsion considered in the Standard Nuclear Model, in the order of 100 times stronger than the electromagnetic interaction.

5) The equilibrium of the newborn nucleus formed by the proton+(original nucleus) is promoted via the equilibrium between the action of the centrifugal force trying to expel the protons and neutrons of the newborn nucleus and the magnetic force actuating on the protons.

6) The stability of the light nuclei via equilibrium between magnetic force and centrifugal force is shown and calculated in my paper Stability of Light Nuclei, published in the JoNP:

http://www.journal-of-nuclear-physics.com/files/Stability%20of%20light%20nuclei.pdf

As the nuclear theorists are now accepting the reality of the cold fusion, thanks to the performance of the E-Cat, sure that they will realize that there is need a new nuclear model for explaining the results obtained by Andrea Rossi.

And as the results obtained by the E-Cat are pointing out that the strong nuclear force cannot be the responsible for the agglutination of the nuclei, then facing the question on what will be the new nuclear model to be chosen (between the many new nuclear models proposing the electromagnetism as the cause of agglutination of the nuclei) of course they will consider only those models capable to explain the puzzle:

how can the electromagnetism to promote the agglutination of the nuclei, since there is need a force 100 times stronger than that promoted by the electromagnetism?

The best new nuclear model able to solve the puzzle will be chosen.

Regards

wlad

The fundamental cold fusion mechanism

• Wladimir Guglinski

November 21st, 2014 at 7:07 PM

Andrea Calaon wrote in November 20th, 2014 at 7:33 AM :

''Let me spend a few words to advertise my theory .

In the reactions above I explicitly added (+e) because I believe that the “secret” of the LENR is a coupling with the electron. Li7 in a uncommon “physical-chemistry” event in the metal matrix, couples with one electron becoming a sort of “new particle”: Li7e. Then this pseudo-particle, which is neutral already at picometric scales, can easily couple (through the same mechanism) with a Ni isotope: Li7eNixx. Li7 and Nixx become forced to travel inside the “circular” electron potential well. Soon they reach “nuclear contact” (at 2-3 [fm]) with very low excess kinetic energy, and can exchange the neutron because it is energetically convenient and probably Li7 offers it on the plane orthogonal to its magnetic moment, right where Nixx can easily “grab it”. ''

—————————————————————————

Dear Calaon,

I have analysed your idea on the “new particle” Li7e, taking in consideration my nuclear model, and I have arrived to some interesting conclusions.

Let me explain it.

Figure 1 ahead shows the nucleus 2He4 with its positive electric field, produced by the two protons.

FIG. 1:

Image:Calaon-guglinski-FIGURE1.png

The nucleus 2He4 has spin about the z-axis shown in the Figure 1.

However, the two protons have residual repulsion (not in that magnitude of the repulsion considered in the Standard Model, because the electric fields of the protons are immersed within the electric field of the 2He4), and due to the repulsion the two protons have oscillations (zig-zag motion), and since the neutrons are bound to the protons via the spin-interaction, the neutrons also oscillate.

Due to the oscillation of the two protons and two neutrons, the z-axis is changing its direction all the time. By this reason in average the positive field of the 2He4 is spherical, as shown in the Figure 2, and the two electrons in the electrosphere take the levels s1 and s2.

FIG. 2:

Image:Calaon-guglinski-FIGURE2.png

Now consider the 3Li7 nucleus, shown in the Figure 3.

FIG. 3:

Image:Calaon-guglinski-FIGURE3.png

The magnetic field of the 3Li7 is shown by North-South (blue-pink).

The magnetic force which links the deuteron to the central 2He4 is induced by the rotation of the proton. The neutron has no charge, and therefore it does not induce magnetic force. The centrifugal force tries to expel the neutron, but it is bound to the deuteron due to spin-interaction.

The radius of the orbit of the deuterion is 0,405fm, while the radius of the orbit of the neutron is 2,391fm. The two values are calculated in the paper Stability of Light Nuclei published in JoNP, based on the equilibrium between magnetic force on the proton and the centrifugal force on the deuteron-neutron, and I had used the magnetic moment of the 3Li7 measured in the experiment so that to calculate the values 0,405 and 2,391.

http://www.journal-of-nuclear-physics.com/files/Stability%20of%20light%20nuclei.pdf

As the neutron in the 3Li7 is bound to the deuteron via the spin-interaction, but the radius orbit of the neutron is very big (2,391fm), it means that the neutron is weakly bound to the deuteron (and it is the deuteron that avoids the neutron to be expeled by the action of the centrifugal force).

As happened in the case of the 2He4, the three protons of the 3Li7 are submitted to oscillations due to repulsions, and as the neutron is bound to the deuteron, also the neutron has oscillation.

So, in spite of the deuteron-neutron move about the z-axis, however the z-axis has a chaotic motion, changing its direction all the time.

Therefore, in average the positive electric field of the 3Li7 due to the three protons is spherical, and the distribution of the electrons in the positive electrosphere of the Li7 takes the levels shown in the Figure 4.

FIG. 4:

Image:Calaon-guglinski-FIGURE4.png

In the page 3 of the Lugano Report it is said:

”Three braided high-temperature grade Inconel cables exit from each of the two caps: these are the resistors wound in parallel non-overlapping coils inside the reactor.”

Therefore the electric current in the coils induces an internal magnetic field inside the alumina cylinder of the reactor, and when a nucleus 3Li7 approaches to a nucleus 28Ni58 and they couple chemically, both the Li7 and the Ni58 align their nuclear z-axis toward the axis of the alumina cylinder of the E-Cat.

Being the two z-axis of both Li7 and 58Ni aligned toward the same direction, the two nuclei couple their nuclear magnetic moment, and by this way both the nuclei of 3Li7 and 58Ni stop to gyrate chaotically, and so the nuclear z-axis of the 3Li7 and 58Ni stops of changing their direction: their nuclear z-axis keep the same direction of the axis of the alumina cylinder.

As the two nuclei stopped to gyrate chaotically, then the two positive electrospheres of 7Li and 58Ni lose the spherical shape they had when they were gyrating chaotically. In other words, both nuclei of 7Li and 58Ni get back the shape of electrosphere shown in the Figure 1 for He4 and Figure 3 for Li7.

This changing in the electrosphere of the 7Li is shown in the Figure 5, where we see that the electrons s1, s2, and p1 change their orbits.

FIG. 5:

Image:Calaon-guglinski-FIGURE5.png

But note that the electron of the level p1 occupies an unstable level, because its negative charge is attracted not only by the positive electrosphere of the Li7, but it is also attracted by the positive electrosphere of the Ni58.

So, the electron of the level p1 is attracted by the electrosphere of the Ni58, and then the electron p´1 changes its orbit, taking the place shown in the Figure 6, between the nuclei Li7 and Ni58.

FIG. 6:

Image:Calaon-guglinski-FIGURE6.png

The orbit of the electron of the level p1 works now like a coil inducing a strong magnetic moment toward the direction of the two nuclear z-axis of the two nuclei 7Li and 58Ni.

As the neutron in the 7Li is weakly bound, and it has a big magnetic moment (-1,913), it will be pulled by the magnetic field of the electron p1 toward the direction of the nucleus Ni58.

Due to the inertia, the neutron continues moving, and it enters within the Ni58 through the “hole” in the electrosphere of the Ni58.

NOTE: look at the Figure 3 of the paper Stability of Light Nuclei the magnetic moment of the neutron within the nuclei depends on the position of the neutron. When the neutron is crossed by a flux-n(o) down being in the outer side of DOUGLAS, its magnetic moment becomes positive: +1,913.

FIG. 3 of the paper Stability of Light Nuclei ( http://www.journal-of-nuclear-physics.com/files/Stability%20of%20light%20nuclei.pdf ):

Image:Fig. 3.JPG

Therefore, the 7Li transmutes to 6Li, and 58Ni transmutes to 59Ni.

The same happens with the isotopes 60Ni, 61Ni, 62Ni.

I think your theory has chance to be correct, dear Calaon. But it seems there is no way to conciliate your theory with the Standard Model.

I think your theory requires my nuclear model so that to explain the Rossi’s Effect.

Regards

Wlad

Comment by Eernie1

November 22nd, 2014 at 12:09 PM

Dear Andrea C and Wlad.

Fermi, Alvarez and Wick have shown both theoretically and experimentally that the injection of an electron into the nucleus occurs naturally in some of the heavier atoms, causing them to be radioactive, emitting Beta particles. These electrons(usually s or p level) are considered present in the nucleus either field wise or particle wise dependent upon whether the investigator treats them as particles or an EM field. The process is called Reverse Beta, conversion electrons, or just Beta decay since the electron presence is subsequently ejected form the nucleus along with a Beta+ or Beta- particle, a Neutrino and a photon whose energy depends on the angle of entrance of the electron. Once the influence of the electron is felt in the nucleus, its spin and its field energy can play havoc with the equilibrium of the resting nucleus resulting in perhaps some strange outcomes. The inner electrons can also be induced to enter the nucleus by imposing an outer negative field on the electron sphere(perhaps a negative Hydrogen ion?).

Energetic regards.

Reply by Guglinski

November 22nd, 2014 at 7:44 PM

Dear Eernie,

it is not the case.

Fermi, Alvarez, and Wick were speaking about absorption of electrons of the inner levels s and p of some heavy nuclei by those own nuclei.

In the case of the Rossi’s Effect, Calaon and I are not speaking about the electrons of the inners levels s and p of the heavier nucleus Ni being absorbed by the own Ni.

We are speaking about the contribution of the inner levels s and p of the lighter 7Li in the transmutation of the heavier Ni.

As you know, dear Eernie, in 2006 was published my Quantum Ring Theory where I had predicted that even-even nuclei with Z=N have non-spherical shape. People used to call me mad, because I had the audacity of defy a dogma considered untouchable along 80 years by the nuclear theorists, according to which those nuclei have spherical shape.

But in 2012 the journal Nature published a paper showing that my prediction was correct: even-even nuclei with Z=N have non-spherical shape:

http://www.nature.com/nature/journal/v487/n7407/full/nature11246.html

More recently, since 1989 the nuclear theorists have considered along more than 25 years that cold fusion is impossible.

They supposed cold fusion to be impossible because according to the Standard Nuclear Physics the positive electrosphere of the nuclei has spherical shape.

So, they believed that, if a particle positively charged will enter within a nuclei, it must win the Coloumb barrier under conditions of high conditions of pressure and temperature (hot fusion), because the Coulomb barrier involves spherically the whole nucleus.

But they are wrong.

The shape of the positive electric field of the nuclei is non-spherical, as the nuclear theorists believed along 80 years.

However, due to the chaotic rotation of the nuclei, in average the positive electric field of the nuclei takes the spherical shape. And this is the reason why hot fusion occurs needs to occur in the Sun.

Soon or later, the nuclear theorists will realize that Rossi’s Effect must be explained via the consideration that the positive electrosphere of the nuclei is non-spherical, and this nuclear property is responsible for the occurrence of cold fusion.

And soon or later, the nuclear theorists will realize that, again, I am right.

Eernie,

perhaps we are witnessing the birth of a new theory, to be known in the future as Calaon-Guglinski theory.

regards

wlad

Question by Joe

November 22nd, 2014 at 11:37 PM

Wladimir,

1. Why do you have only the outer electron of 3Li7 involved in the process of neutron transfer? Why are the outer (3d, 4s) electrons of the 28NiXX not involved at all in the Calaon-Guglinski neutron transfer process?

2. Why would the valence neutron at 7fm prefer exiting along the z-axis in which direction it has no momentum than along the xy-plane in which it has angular momentum? (Remember that Andrea Calaon has the two nuclei with their z-axes parallel rather than collinear as per your view.)

All the best,

Joe

Reply by Guglinski

November 23rd, 2014 at 7:31 AM

Joe wrote in November 22nd, 2014 at 11:37 PM:

1. ) ———————————–--------------------------

Why do you have only the outer electron of 3Li7 involved in the process of neutron transfer? Why are the outer (3d, 4s) electrons of the 28NiXX not involved at all in the Calaon-Guglinski neutron transfer process?

-------------------------------------------

Joe,

perhaps they are also involved, since the Ni also changed the shape of its positive field due to the nucleus, and so the electrons in the electrosphere change their orbits, and the outer electrons of the Ni have interaction with the positive field of the 7Li.

However, in order to simplify the explanation, I had explained only what happens with the outer electron of the 7Li.

2. ) ——————————–---------------------------

Why would the valence neutron at 7fm prefer exiting along the z-axis in which direction it has no momentum than along the xy-plane in which it has angular momentum?

--------------------------------------------

First of all, the neutron is not at 7fm, actually it is at a distance of 2,391fm.

The neutron exits along the z-axis because the orbit of the electron p1 in the Figure 6 induces a magnetic dipole moment vector along the z-axis:

http://en.wikipedia.org/wiki/Magnetic_moment

FIG. 6:

Image:Calaon-guglinski-FIGURE6.png

As the magnetic moment vector of the neutron is also along the z-axis, then the neutron is pulled by the magnetic moment of the electron p1 toward the z-axis.

3. ) ———————————–-----------------------------

(Remember that Andrea Calaon has the two nuclei with their z-axes parallel rather than collinear as per your view.)

----------------------------------------------

Calaon has not a new nuclear model so that to allow him to understand the physical mechanism involved in the Rossi’s Effect.

Note that he even did not respond to my question, when I asked to him how to solve that puzzle regarding the nuclear models which do not consider the strong force as the cause of the agglutination of the nuclei: as the electromagentism is 100 times weaker than the Coulomb repulsions in the distances of 2fm within the nuclei, how can the electromagnetis to be responsible for the nuclei aggregation?

So, he is trying to understand what happens (as everybody) by considering what he knows from the known models (in which the electrosphere of the nuclei is spherical and unalterable).

Then Calaon needs to decide what he prefers to do.

He can either keep his initial version or to adopt the new way I am suggesting to him.

regards

wlad

Comment by Andrea Calaon

November 23rd, 2014 at 5:49 PM

Dear Wladimir Guglinsky,

As you probably noticed, I have already “withdrawn” the idea of an actual coupling between Li nucleus and an electron, because it seems to me impossible that the coupling can cross the two 1s electrons protecting the Li+ ion.

I have to premise that I haven’t analysed your theory in detail. But there are some features I don’t understand or probably just do not agree with.

You seem to say that the shape of the “electrostatic” field (intensity in different directions) is influenced by the kinetics of the particles. Let me say that, if I understood correctly, I can not agree, because the electric field symmetry is a basic feature of electromagnetism. As a consequence the so called Coulomb barrier is identical in all directions.

My “theory”, which is Dallacasa’s in this respect, explains the “non-sphericity” of the nuclear attractive force among nucleons with the strong dependence of the attractive potential with the reciprocal orientation of the magnetic moments (i.e. spins) (and their phasing). Nothing exotic, just basic electromagnetism.

You say: “When the neutron is crossed by a flux-n(o) down being in the outer side of DOUGLAS, its magnetic moment becomes positive: +1,913.” I don’t understand what you are saying. Sorry.

I am trying to follow the suggestion of our Italian “bon-ton Guru” orsobubu putting a bit more “pepper” in the discussion.

But you know that actually I would never criticize someone’s work without a reason, and would never offend (consciously) anyone for having a different opinion.

Wladimir, let me insist that what I proposed is not going against any consolidated law of physics, not to mention the so called Standard Model. I am sure that there is no need to contradict any main evidence of physics to explain LENR. So far I just took the possible validity of the nuclear potential of Dallacasa to its extreme consequences.

Regards to all

Andrea Calaon

Reply by Guglinski to Calaon

November 23rd, 2014 at 8:24 PM

Andrea Calaon wrote in November 23rd, 2014 at 5:49 PM

1) ———————————————--------------------------------

You seem to say that the shape of the “electrostatic” field (intensity in different directions) is influenced by the kinetics of the particles. Let me say that, if I understood correctly, I can not agree, because the electric field symmetry is a basic feature of electromagnetism. As a consequence the so called Coulomb barrier is identical in all directions.

---------------------------------------------------

No, you did not understand.

The field of particles and the field of the nuclei is composed by two spherical fields.

Look for instance the two fields of the 2He4:

Image:Calaon-guglinski-FIGURE1.png

But due to the chaotic rotation the fields takes in average the spherical shape:

Image:Calaon-guglinski-FIGURE2.png

This explains why the Coulomb barrier is identical in all directions

2) ————————————–----------------------------

You say: “When the neutron is crossed by a flux-n(o) down being in the outer side of DOUGLAS, its magnetic moment becomes positive: +1,913.” I don’t understand what you are saying. Sorry.

---------------------------------------------

My theory is developed from the concept of non-empty space (aether), composed by elementary particles of the aether as electricitons, magnetons, gravitons, permeabilitons.

The concept of field in my theory emeges from the formation of physical fields composed by electricitons, magnetons, gravitons, and permeabilitons.

All the nuclei are composed by a central 2He4, which produces a gravity flux named flux-n(o). Each particle as proton and neutron is captured by such flux.

Due to the laws of interactions between the gravitons of the flux-n(o) and the spin of particles (protons and neutrons), the magnetic moment of those particles can change their sign depending on the direction of the flux-n(o).

If a flux-n(o)-up crosses the neutron, its magnetic moment is positive.

If a flux-n(o)-down crosses the neutron, its magnetic moment becomes negative.

3) ——————————--------------------------------

Wladimir, let me insist that what I proposed is not going against any consolidated law of physics, not to mention the so called Standard Model.

---------------------------------------------

According to the Standar Model nuclear reactions cannot occur via electromagnetism, and the resaon is easy to be understood: nuclear reactions need to be promoted by particles bound via strong nuclear force.

Your theory requires a model in which protons and neutrons are bound via electromagnetism.

Therefore your theory is going against the Standard Model.

As I already said , the Coulomb barrier in the distances of 2fm within the nuclei is 100 times stronger than the electromagnetism interaction.

There is need a new nuclear model so that to explain how protons and neutrons can be bound via electromagnetism.

The Lugano Report is showing that the results obtained from Rossi’s E-Cat are incompatible with nuclear reactions occuring via strong force, as you did point out.

Therefore Rossi’s Effect is incompatible with the Standard Model, based on the hypothesis of protons and neutrons bound via strong force within the nuclei.

regards

wlad

Second reply by Guglinski to Calaon:

November 24th, 2014 at 3:26 AM

Andrea Calaon wrote in November 23rd, 2014 at 5:49 PM:

————————————–-----------------------------------------

You say: “When the neutron is crossed by a flux-n(o) down being in the outer side of DOUGLAS, its magnetic moment becomes positive: +1,913.” I don’t understand what you are saying. Sorry.

------------------------------------------------------

With figures is easier to understand.

The flux-n(o) of the 2He4 is shown in the figure:

Image:Fig. 3.JPG

In the inner side of DOUGLAS the neutron has magnetic moment -1,913 , because it is crossed by a flux-n(o)-up.

The neutron in the outer side of DOUGLAS has magnetic moment +1,913, because it is crossed by a fluz-n(o)-down.

All the other nuclei are formed by the capture of deuterons and neutrons by the flux-n(o) of the 2He4.

In the figure ahead the 3Li7 is formed by the central 2He4 and the deuteron-neutron captured by the flux-n(o)

Image:Calaon-guglinski-FIGURE3.png

The positive field of the proton is similar to the positive field of the 2He4:

Image:Calaon-guglinski-FIGURE1.png

The field is non-spherical

But as the proton is made by quarks, and there is repulsion between the quarks up, then the body of the proton has chaotic spin, then in average the field of the proton becomes spherical.

regards

wlad

A good question, by Joe

November 24th, 2014 at 3:32 AM

Wladimir,

If the induced magnetic dipole of the rotating 1p1 electron of 3Li7 can attract the valence neutron of 3Li7, why would that neutron not stop and settle at the centre point of the rotation rather than proceed further and into the 28NiXX nucleus?

All the best,

Joe

Reply by Guglinski

November 24th, 2014 at 8:13 AM

As I said in my comment of November 21st, 2014 at 7:07 PM:

“Due to the inertia, the neutron continues moving, and it enters within the Ni58 through the “hole” in the electrosphere of the Ni58.”

regards

wlad

On the non-spherical Coulomb barrier of the nuclei

• Wladimir Guglinski

November 25th, 2014 at 8:13 AM

Dears Joe, Calaon, Orsobubu, Karrels, Eric…

… and anybody interested in the subject.

Dr. Stoyan Sarg is going to pronounce a speech where he says that Coulomb barrier was wrongly interpreted in scattering experiments:

“At the beginning it discuses the major methodological error in scattering experiments that leads to a tremendously wrong vision about the Coulomb barrier.”

http://www.e-catworld.com/2014/11/21/dr-stoyan-sarg-to-address-nanotek-expo-2014-on-lenr/

But Dr. Sarg is wrong.

There is nothing wrong with the scattering experiments, and the vision about the Coulomb barrier is correct.

In normal condictions (different of those occurring in cold fusion experiments) the electric field of nuclei (Coulomb barrier) is spherical, as correctly interpreted by the nuclear theorists, because of the following:

1- The nuclei have non-spherical Coulomb barrier.

For instance, the figure shows the Coulomb barrier for the 2Her, shown as yellow in the figure.

FIG. 1:

Image:Calaon-guglinski-FIGURE1.png

2- But the nuclei have chaotic rotation (due to repulsions between protons) and the z-axis of the Figure 1 is changing its direction every time.

3- As consequence of the chaotic rotation, the electric field takes in average the spherical shape, as shown in the Figure 2 ahead for the 2He4.

FIG 2:

Image:Calaon-guglinski-FIGURE2.png

4- The spherical Coulomb barrier of the Figure 2 was detected in the scattering experiments, and so the vision of the Coulomb barrier by the physicists was correct

5- However, in cold fusion phenomena occurs the alignment of the two z-axes of two nuclei (as for instance 7Li and 58Ni in the Rossi’s Effect).

We see it in the Figure 5 ahead.

FIG 5:

Image:Calaon-guglinski-FIGURE5.png

6- The two Coulomb barriers of 7Li and 58Ni take their original non-spherical shape in the Figure 5 because the z-axes of the two nuclei stop to gyrate chaotically. This happens only in the cold fusion experiments (for instance, in the Rossi’s E-Cat the z-axes of 7Li and 58Ni are aligned along the axis of the alumina cylinder, because of the magnetic field created by the electric current in the coils).

7- As we realize from Figure 1, there is a “hole” in the Coubomb barrier of the nuclei, along the z-axis:

FIG 1:

Image:Calaon-guglinski-FIGURE1.png

8- When the two z-axes of two nuclei are aligned (as 7Li and 58Ni in the Figure 5), the two “holes” of the two nuclei are aligned, and so it is easier for a particle as a proton or a neutron to exit one of them and to enter within the other.

regards

wlad

On the interaction between the neutron and the electron orbit

Eric Ashworth

November 25th, 2014 at 10:44 PM

Wladimir, Your reply to Joe Nov 24th 2014. “As I said in my comment of Nov 21st. 2014. “Due to the inertia, the neutron continues moving and it enters within the Ni58 through the “hole’ in the electrosphere of the Ni58?.

This reply of yours I believe to be correct but you use the word inertia which I do not think is fully understood as a cause of an effect. For me inertia is a cause of movement but what causes the object to continue to move when the propelling force is removed?. There can also be an attractive force.

Reply by Guglinski to Eric

November 26th, 2014 at 9:23 AM

Dear Eric, in absence of force there is conservation of momentum P= m.V . The inertia does not depend on actuation of a propelling force.

However, in the case of the neutron, when the neutrons is moving toward the Ni nucleus, it can happen the following:

1) Before to arrive to the cross-section of the electron’s orbit, the magnetic field induced by the electron’s motion applies a force of ATTRACTION on the neutron (the two magnetic vectors point to the the same direction).

The neutron is PULLED by the electron’s orbit toward the Ni nucleus.

2) After crossing the cross-section, the magnetic field of the electron starts to apply a force of REPULSION on the neutron (because the two magnetic vectors continue with the same direction).

And so the neutron is PUSHED toward the Ni nucleus.

However, I did not mention it because I am not sure if the neutron changes its magnetic field regarding the electron’s orbit after crossing the cross-section of the electron’s orbit (in this case, if the neutron changes the magnetic field vector in the contrary direction, then the electron begins to apply a force of ATTRACTION on the neutron, after it crosses the cross-section, and therefore decreasing the speed of the neutron going by inertia toward the Ni nucleus).

regards

wlad

Question by Joe

November 27th, 2014 at 1:20 AM

Wladimir,

Since your model must necessarily involve a target nucleus that has a greater magnetic dipole moment than the source nucleus in order that a nucleon be successfully drawn, compare the relative strengths of the magnetic dipole moments of Fe and Co and of Ca and Sc.

All the best,

Joe

Reply by Guglinski

November 27th, 2014 at 9:13 AM

Not necessarily, Joe

7Li magnetic moment is greater than Ni.

For instance, 7Li has magnetic moment +3,25, while 28Ni61 has magnetic moment -0,75

What pulles the neutron of the 7Li toward the Ni is the orbit of the electron p1 (shared by 7Li and Ni), which magnetic moment is very greater than that of the 7Li:

http://peswiki.com/index.php/Image:Calaon-guglinski-FIGURE6.png

regards

wlad

Question by Joe

November 27th, 2014 at 6:03 PM

Wladimir,

Every nucleus that you mentioned has a magnetic moment that is smaller than that of 3Li7. If the 1p1 electron of 3Li7 is able to pull a nucleon away from the nucleus of 3Li7, it should be even easier to pull nucleons out of the target nuclei that you mentioned. The result would be a collision of nucleons from both nuclei – intended source and intended target – right at the centre of rotation of the 1p1 electron.

All the best,

Joe

Reply by Guglinski

November 27th, 2014 at 6:55 PM

Joe, two things:

1- 7Li has a neutron weakly bound

2- 7Li has 3 protons, its electrosphere is small, compared with the electrosphere of Ni, which has 28 protons.

Therefore the orbit of the electron p1 in the Fig. 6 is very nearest to the nucleus 7Li and far away from the nucleus Ni, As the magnetic force decreases with the square of the distance, you may realize that the magnetic force on the neutron of 7Li is very stronger than on the neutrons of the Ni.

http://peswiki.com/index.php/Image:Calaon-guglinski-FIGURE6.png

The neutron (or proton) will always exit from the lighter nucleus to the heavier one.

regards

wlad

Question by Joe

November 28th, 2014 at 7:30 PM

Wladimir,

Problem 1.

If a very strong magnetic moment can pull a nucleon from a nucleus with a fairly strong magnetic moment, why would the nucleon stay in the target nucleus which has a weak magnetic moment? The factor of inertia is a non-issue. The same force that pulled the nucleon out of the source nucleus will have a much easier time to stop the nucleon’s progress toward the target nucleus when the nucleon crosses the plane of rotation of the 1p1 electron of 3Li7. And when we add to this the matter of distance, like you mentioned, which makes the attraction to the target nucleus even weaker for the nucleon, then I do not see much success for this model.

Problem 2.

Even if Problem 1 did not exist, there exists the matter of the outer electrons of 28NiXX having a role to play which they do not presently have in your model. Those outer electrons might also rotate and pull nucleons out of 28NiXX, especially if they are more loosely bound than the 1p1 electron of 3Li7. (Mind you, it is also possible that they rotate in such a way as to pull nucleons INTO 28NiXX which would then help make your model viable.)

All the best,

Joe

Comment by Guglinski on the magnetic moments for 58Ni and 60Ni

November 28th, 2014 at 10:34 PM

Dears Joe and Steven Karels

58Ni and 60Ni have magnetic moment zero

http://www.webelements.com/nickel/isotopes.html

Therefore, the oscillatory magnetic field applied within the reactor is used not only for shaking the nuclei, but also for exciting them, because a nucleus with magnetic moment zero cannot be aligned by a magnetic field

https://www.google.com.br/search?q=nucleus+excitation&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:pt-BR:official&client=firefox-a&channel=sb&gfe_rd=cr&ei=WzF5VIi6E43dwAT_rIKQDw

Excited 58Ni has magnetic moment -0,1, while excited 60Ni has magnetic moment +0,2

regards

wlad

Reply to Joe by Guglinski

November 28th, 2014 at 11:07 PM

Joe wrote in November 28th, 2014 at 7:30 PM:

Problem 1.

If a very strong magnetic moment can pull a nucleon from a nucleus with a fairly strong magnetic moment, why would the nucleon stay in the target nucleus which has a weak magnetic moment? The factor of inertia is a non-issue.

------------------------------------------------

Dear Joe,

I already explained it to Eric:

————————————-----------------------------------

However, in the case of the neutron, when the neutrons is moving toward the Ni nucleus, it can happen the following:

1) Before to arrive to the cross-section of the electron’s orbit, the magnetic field induced by the electron’s motion applies a force of ATTRACTION on the neutron (the two magnetic vectors point to the the same direction).

The neutron is PULLED by the electron’s orbit toward the Ni nucleus.

2) After crossing the cross-section, the magnetic field of the electron starts to apply a force of REPULSION on the neutron (because the two magnetic vectors continue with the same direction).

And so the neutron is PUSHED toward the Ni nucleus.

However, I did not mention it because I am not sure if the neutron changes its magnetic field regarding the electron’s orbit after crossing the cross-section of the electron’s orbit (in this case, if the neutron changes the magnetic field vector in the contrary direction, then the electron begins to apply a force of ATTRACTION on the neutron, after it crosses the cross-section, and therefore decreasing the speed of the neutron going by inertia toward the Ni nucleus).

———————————-----------------------------------------------------

Problem 2.

Even if Problem 1 did not exist, there exists the matter of the outer electrons of 28NiXX having a role to play which they do not presently have in your model. Those outer electrons might also rotate and pull nucleons out of 28NiXX, especially if they are more loosely bound than the 1p1 electron of 3Li7. (Mind you, it is also possible that they rotate in such a way as to pull nucleons INTO 28NiXX which would then help make your model viable.)

-----------------------------------------------------------------

Joe,

due to Pauli’s Principle, only one electron can take the orbit taken by the electro p1 of the 7Li in the Figure 6 (orbit perpendicular to the z-axis)

FIG 6:

Image:Calaon-guglinski-FIGURE6.png

As the inner electron p1 of 7Li is more energetic than the outer electrons of the 58Ni, that orbit in the FIG 6 is taken by the electron of 7Li

So, the electrons of the 58Ni do NOT take an orbit PERPENDICULAR to the z-axis, and they actually cancell each other their magnetic moments (the vector magnetic moment for the electron p1 has contrary direction of the vector for the electron p2, and the vector for the electron d1 has contrary direction of vector for the electron d2, etc)

regards

wlad

Fleischmann-Pons and Pamela Mosier-Boss experiments finally deciphered

• Wladimir Guglinski

November 29th, 2014 at 12:10 AM

Fleischmann-Pons experiment and Mosier-Boss experiment are practically the same.

But Mosier-Boss experiment has a special interest, because in her experiment he measured the energy of the neutrons emitted, in order of 10MeV, and such result is paradoxical, since the binding energy of the deuterons is only 2,2MeV.

Let us see the mechanism of the cold fusion reactions in the Mosier-Boss experiment.

1) A Pd nucleus couples with a deuteron, and they align their z-axes.

Therefore both the Pd and 1H2 lose their spherical Coulomb barrier.

2) The electron s1 of the deuteron takes an orbit perpendicular to the z-axes of the two nuclei Pd and 1H2

3) The deuteron has only one proton, and so its positive electric field is small. Therefore the radius orbit of the electron s1 is short. By consequence, the magnetic force produced by the s1 orbit is not able to remove the neutron tied to the proton in the deuteron.

4) The couple Pd-1H2 is stable, and nothing happens.

5) Then the couple Pd-1H2 capture a second deuteron, and then a sandwish is made: Pd-1H2-1H2

6) The electron s1 of the second deuteron takes an orbit perpendicular to the z-axis, between the two deuterons.

7) Therefore the first proton-neutron of the first deuteron (sandwihed by Pd and the second deuteron) start to have a high oscilatory motion, because:

7.a) The first deuteron is between two parallel orbits, one orbit of the electron s1 of the first deuterion, and other orbit of the electron s1 of the second deuteron

7.b) the positive charge of the proton of the first deuteron is attracted by two contrary directions along the z-axis direction

8) The neutron is tied to the proton in the first deuteron by the spin-interaction, which binding energy is 2,2MeV. When, in the oscillatory motion, the proton stops and changes the direction of its motion in contrary direction, due to the inertia the neutron has the tendency to continue its motion, and so the 2,2MeV binding energy is broken, and the neutron exits the first deuteron, and is captured by the second deuteron, and they form a tritium.

9) The capture of the neutron causes an unbalance of masses in the newborn tritium. So, the z-axis of the tritium starts to gyrate chaotically, and it gets back its spherical Coulomb barrier, and it leaves away the couple Pd-1H2-1H3

10) But in the instant when the neutron left away the first deuteron, at that the same time the proton is accelerated toward the contrary directionBut in the instant when the neutron left out The proton gets a high velocity, because free of the link with the neutron the proton is pushed by the repulsion with the positive electric field of the tritium.

11) The proton enters within the Pd nucleus with 2,2MeV, helped by the orbit of the electron s1 pulling it against the Pd nucleus.

12) Within the electrosphere of the Pd nucleus the proton captures an electron, and they form a neutron. That electron in the Pd electrosphere was moving with helical trajectory. As the electron loses its helical trajectory when it is captured by a proton and they form a neutron, the energy of the helical trajectory is transferred to the neutron. So, the neutron gets a portion of the energy 10MeV from the helical trajectory of the electron captured by the proton.

13) When the proton entered within the Pd electrosphere, it was accelerated due to its attraction with the electrons of the Pd nucleus. If the proton was to leave away the Pd electrosphere, it would leave it with the same 2,2MeV, because leaving it its velocity would be decreasing by the attraction with the electrons. However, when the proton becomes a neutron, it loses its charge, and when the neutron starts to leave away the Pd nucleus it is not decelerated by attractions with electrons of the Pd nucleus. Therefore the neutron receives the other portion of the energy 10MeV from the acceleration of the proton before to become a neutron.

Other reactions in Fleischmann-Pons and Mosier-Boss experiments

After the formation of tritium by the first sandwich Pd-1H2-1H2, other sandwiches can be formed, as follows:

1) Pd-1H2-1H3

2) Pd-1H3-1H2

3) Pd-1H3-1H3

In these sandwiches 2He4 is formed

As it is not used an oscillatory magnetic field so that to excite the nuclei, 2He4 is never captured so that to forma sandwich, because 2He4 has no magnetic moment

Pd is not transmutted, because as the proton has charge, it is deviated by the electrons of the Pd electrosphere, and so the proton never hits the Pd nucleus

In Fleischmman-Pons experiment they did not use an external magnetic field, and so the nuclei were aligned by the Earth’s magnetic field.

In the days of magnetic storms of the Sun they did not succeed to replicate the results.

In Mosier-Boss experiment she used an external magnetic field, and so the nuclei were aligned by that field.

Regarding Rossi’s E-Cat

In Rossi’s E-Cat, probably in the first experiments he applied an oscillatory magnetic field unable to excite the nuclei, and therefore only Ni61 had transmutation, because it has magnetic moment -0,75.

But 58Ni and 60Ni had no transmutation, because they have magnetic moment zero, and so they cannot align their z-axis, and therefore they did not capture 7Li nuclei.

By this way Rossi obtained a very low COP.

By improving the oscillatory magnetic field, Rossi did succeed to excite the nuclei, and by this way 58Ni and 60Ni started to have transmutation, and by this way he got to increase the COP of the E-Cat.

Andrea Rossi could confirm it to us, but as he never gives any information about the reactor, he will tell nothing.

Cold fusion requires nuclei with pair number of protons

• Wladimir Guglinski

November 29th, 2014 at 9:22 AM

Joe,

I think only the nuclei with pair number of protons are suitable to transmute by cold fusion, because they also have pair number of electrons in the electrosphere.

Because of the pair number of electrons, each pair of electrons cancell each other their magnetic moment, and so they do not influence the penetration of the neutron.

While the nuclei with odd number of protons, have odd number of electrons, and therefore they always have one unpaired electron, and its magnetic moment do not allow the penetration of the neutron coming from other nucleus.

For instance, suppose Andrea Rossi had used a fuel composed by 29Cu and 3Li7 in his E-Cat. The neutron after exit the 7Li would not succeed to enter within the 29Cu nucleus, because the magnetic moment due to the orbit of the unpaired do not allow it.

It also seems that the best nuclei to get cold fusion is those ones with biggest nuclear magnetic moment, because they can have their z-axis aligned more quickly, and so the reactions will occur faster, and by this way a higher COP will be obtained.

From this viewpoint, 24Cr would be best than 28Ni.

But 24Cr has 24 protons, and therefore its positive electric field is smaller than that of 28Ni. Therefore the orbit radius of the electron p1 of the 7Li will be shorter (since the orbit of the 7Li is influenced by the electric fields of 24Cr and 7Li working together), and by this reason the magnetic moment of the p1 orbit will not be so efficient to extract the neutron of the 7Li.

In the case of a fuel 46Pd-7Li be used in the E-Cat, the radius orbit of the p1 electron of the 7Li will be too much large, and the neutron will be captured from the 7Li with too much energy (high velocity). Then, instead of being captured by the 46Pd nucleus, the neutron actually crosses the 46Pd without to be captured, and the 46Pd will have not transmutation.

Probably this is the reason why, after trying many elements, the tests made by Rossi have pointed that 28Ni is the best fuel to react with 7Li and to get the higher COP

regards

wlad

Again the question on the neutron pulled by the electron’s orbit

• Joe

November 29th, 2014 at 3:51 PM

Wladimir,

If both vectors, that of the neutron’s magnetic moment and that of the electron orbit’s magnetic moment, always stay in the same direction, there will be attraction at the beginning. But when the neutron is almost halfway across the plane of the electron’s orbit, repulsion will be experienced due to the North of the neutron’s vector approaching the North of the electron’s orbit vector. And even if it would continue successfully past the plane, the neutron will experience the force of attraction when its South is pulled by the North of the electron’s orbit. These two conditions should prevent a successful neutron transfer.

The situation is worse if the neutron’s vector is allowed to change its orientation at any time during its travel. The much stronger magnetic moment of the electron’s orbit will ALWAYS end up PULLING the neutron towards it, therefore effectively preventing any neutron transfer from occurring.

All the best,

Joe

Reply by Guglinski

November 29th, 2014 at 9:35 PM

Joe

what you say makes no sense

If there is force of attraction on the neutron when its spin is in clockwise direction, then if the neutron changes its spin in counter clockwise direction the force cannot continue to be of attraction

There was an error working with the wiki: Code[1]

regards

wlad

Reply by Joe

November 29th, 2014 at 10:54 PM

Wladimir,

In your model, does the neutron ever change the direction of its spin? If so, at what point does it do so in its travel from source nucleus to target nucleus?

All the best,

Joe

Reply by Guglinski

November 30th, 2014 at 8:52 AM

Joe,

we are obliged to conclude that yes, the neutron needs to change its spin.

It changes its spin just after crossing the cross section of the p1 electron orbit.

I think the neutron do it as consequence of the least action principle.

As you know, my model of neutron is formed by proton+electron (the electron moving in orbit about the proton).

The proton has magnetic moment +2,793

The neutron has magnetic moment -1,913

So, the orbit of the electron about the proton yields the magnetic moment -(1,913+2,793) = -4,706.

Before crossing the cross-section of the electron orbit p1, the neutron is moving having its electron moving with a parallel orbit with the orbit p1, and intrinsic spin up.

When the neutron crosses the cross-section, the electron of the neutron changes its intrinsic-spin to down. So, the proton of the neutron is constrained to also change its spin, and therefore the neutron starts to move with a contrary spin (contrary to the spin the neutron had before to cross the orbit p1).

Saying the contribution of the least action principle in other words: it is most confortable for the neutron to change its spin after crossing the cross-section of the orbit p1.

regards

wlad

Reply by Joe

November 30th, 2014 at 4:02 PM

Wladimir,

Are you saying that, before the neutron crosses the plane, the neutron’s electron has an orbit about the proton that is parallel to the orbit of the 1p1 electron and that after the neutron crosses the plane, the neutron’s electron has an orbit about the proton that is anti-parallel to the orbit of the 1p1 electron?

All the best,

Joe

Substitute for 3Li7 and 28Ni in Rossi’s eCat

The fuel used in the eCat is composed by an emitter 7Li of neutrons together with Ni isotopes receptors of neutrons.

The emitter 7Li has three electron orbits:

a) two orbits of electrons s1 and s2, which cancel each other their magnetic fields

b) one orbit of electron p1, perpendicular to the z-axes of 7Li, whose magnetic field extracts the neutron of the nucleus 7Li

FIG. 6:

Image:Calaon-guglinski-FIGURE6.png

The radius of the orbit p1 depends on the element used as receptor.

a) By using a receptor with big quantity of protons, the orbit will be larger, and the force of extraction is stronger.

b) By using a receptor with low quantity of protons, the orbit will be smaller, and the force of extraction is weaker.

1- Beryllium 9Be isotope used as emitter

Beryllium has only one stable isotope, the 4Be9.

Figure 7 shows the orbits s and p.

FIG. 7:

Image:FIGURE-7-substitute3Li7-28Ni.png

The orbits s1 and s2 cancell each other their magnetic moments.

The orbits p1 and p2 also cancell each other.

CONCLUSION: Cold fusion cannot be obtained by using Be as emitter

Any element having

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