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Interpretation of John Arrington’s experiment according to Quantum Ring Theory
Posted on Monday, April 09, 2012 @ 19:38:06 GMT by vlad

Science WGUGLINSKI writes: See the the first figure in the link that describes Arrington’s experiment:

That figure shows the beryllium nucleus, with a nucleon which distance to the central 2He4 is 7fm. In the link they say about the beryllium nucleus:

“The surprise came with the beryllium. Unlike the other nuclei, it has two clusters of nucleons, each resembling a nucleus of an atom of helium-4.”

According to Quantum Ring Theory, all the nuclei with Z>2 have a structure with radius in order of 7fm.

Then why does Arrington says “unlike the other atoms”?
Why beryllium is an exception ?

Let’s see why.

According to QRT, the hexagonal floor begins with the isotopes of 3Li. In the link ahead we see the structure of 3Li6, shown in the page 230 of the book QRT:

The 3Li6 has a nuclear spin about the z-axis (vertically upward in the page 230). As the 3Li6 has an unpaired deuteron, it has a strong trepidation.
Then in Arrington’s experiment, due to such big vibration, the unpaired deuteron appears as a clould about the central 2He4 (the same happens with all the 3Li isotopes).

The next nucleus is the 4Be8.
Look at its structure in the page 230, shown beside the 3Li6.
The two nucleons 1H2 occupy two positins diametrally opposed regarding to the central 2He4, and so the three nucleons form a straight length 14fm.

Looking at the structure of 4Be8, we realize that it has no trepidation, because the masses of the two nucleons 1H2 are distributed symmetrically about the central 2He4 (that’swhy the 4Be8 has null electric quadrupole moment).
Therefore, when the two deuterons gyrate about the z-axis (because of the nuclear spin about the z-axis which crosses the central 2He4), they appears in the Arrington’s experiment as a straight line very well defined, 14fm long.

The next nucleus is the 5B. It has one unpaired deuteron, and so it has a big trepidation. Due tothe nuclear spin, its image in the experiment appears as a cloud about the central 2He4

The next nucleus is 6C. In the detail of page 231 we see the structure of 6C12. Its structre forms a cube with diagonal of about 14fm around the central 2He4. So, in spite the 6C12 has no trepidation, however due to the nuclear spin the four deuterons 1H2 appears in the images of the experiment as a cloud about the central 2He4.

The next nucleus is the 7N. As it has one unpaired deuteron, its image shows in the experiments a cloud about the central 2He4.

The next nucleus is 8O.
The structure of 8O16 is shown in page 144 of the book QRT. The page is shown in the link previously posted.
The structure of 8O16 is not flat. Due to repulsion between the 6 nucleons 1H2, they oscillate about the central 2He4. This is shown in the Fig. 1.2 of the page 144.
So, in the image captured by the Arrington’s experiment, the 8O16 shows a cloud distributed about the central 2He4.

After the nucleus 8O, a new hexagonal floor beggins, with the nucleus 9F. As it has an unpaired deuteron, its image shows a cloud about the central 2He4.

All the other nuclei exhibit a cloud about the central 2He4.
That’s why Arrington’s experiments showed the beryllium as a surprise. Only its image shows a straight line of length 14fm.

That’s why beryllium is an exception.



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EQM of beryllium in Arrington’s experiment contradicts Nuclear Physics (Score: 1)
by vlad on Tuesday, April 10, 2012 @ 21:34:32 GMT
(User Info | Send a Message) http://www.zpenergy.com

WGUGLINSKI writes: See the first figure in the link that describes Arrington’s experiment:
Link 1:

That figure shows the beryllium nucleus, with a nucleon which distance to the central 2He4 is 7fm.

From experiments, it’s known that beryllium 4Be8 has null electric quadrupole moment, Q(b)=0.
But from current Nuclear Physics it’s impossible to explain how the structure detected by Arrington’s experiment may have Q(b) = 0.

Let’s see why.

A spherical distribution of electric charge has Q(b)=0. But looking at the beryllium’s structure detected in Arrington’s experiment we realize that, from the nuclear models of Nuclear Physics, that structure cannot have Q(b)=0, because that distribution of charge in 4Be8 is not spherical, since there are three positive charges aligned along a straigth line, and therefore such structure of 4Be8 could not have Q(b)=0.

Let’see how to explain why 4Be8 has Q(b)=0 by considering the hexagonal floors model of Quantum Ring Theory.

The structure of 8O16 is shown in the page 144 of the book QRT, shown in the link:
Link 2:

The nucleus 8O16 has Q(b)=0 , and its nuclear magnetic moment is also null.

If the hexagonal floor formed by six nucleons 1H2 around the central 2He4 in the 8O16 was flat, it could not have Q(b)=0, because its charge distribution would not be spherical.

But look at in the link 2 the detail of page 144 showing the Fig. 1.2, where we see what happens with the nucleons 1H2 of the hexagonal floor of 8O16:
a) the nucleons 1H2 have oscillation
b) in the SIDE-VIEW of that figure we see that they oscillate about the x-y plane (which is orthogonal to the z-axis, about which the nucleus gyrates, performing its nuclear spin).
c) so, due to the oscillation of the six deuterons 1H2, the z-axis is changing continously its direction, chaotically.

As the nucleus 8O16 has null nuclear magnetic moment, there is no way to get its alligment along a direction, by applying a strong external magnetic field. So, even within a strong external magnetic field to be used in experiments, the z-axis of the nucleus 8O16 continues always changing its direction.

Therefore, in average, the nucleus 8O16 behaves as if should have a spherical distribution of charge. In another words: when the researchers measure the electric quadrupole of 8O16 in their experiments, they get Q(b) = 0 because in average its distribution of charge behaves as a spherical distribution.

This is explained in the book Quantum Ring Theory, in the chapter “Electric Quadrupole Moment”, at the page 136, where it is also shown why the isotope 8O18 has an anomaly that cannot be explained from the models of Nuclear Physics.

The same happens with the nucleus 4Be8. It has null nuclear magnetic moment. It’s easy to realize why, by looking at the structure of 4Be8 shown in the page 230 of QRT (see the link 2), because:
a) the central 2He4 has null nuclear magnetic moment
b) the two deuterons 1H2 gyrate about the z-axis, and each magnetic moment of them cancell one each other.
c) the two deuterons 1H2 also have oscillation, due to repulsion with the central 2He4.

Therefore, when the researchers measure the electric quadrupole moment of 4Be8 by experiments, they get Q(b) =0, because the z-axis changes its direction everytime, (since they cannot align the nucleus 4Be8 by applying a strong external magnetic field) and so in average its distribution of charge takes the spherical shape.

As we realize, John Arrington’s experiment contradicts the models of Nuclear Physics in many aspects.

Finally, we have to consider the following:
1- In general, the quantum theorists reject to consider a New Physics for explaining cold fusion because they want to keep their current Theoretical Physics.
2- However, we realize that, even if we do not consider the cold fusion experiments, there is no way to keep the current Physics, because there are many other experiments which result are requiring a New Physics, and the Arrington’s experiment is one among them.
3- Therefore, as there is no way to avoid to start to take in consideration a New Physics, it makes no sense such their attempt of trying to keep the current Physics for explaining cold fusion. It is not cold fusion, itself, which requires a New Physics. It is, in general, several new experiments which are requiring it.


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