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Why a new physics theory could rewrite the textbooks
Posted on Friday, January 29, 2016 @ 12:56:13 EST by vlad

Science WGUGLINSKI writes: To: Prof. Anthony W. Thomas, Australian Research Council Laureate Fellow, University of Adelaide

Dear Prof. Anthony W. Thomas

Regarding the paper “Why a new physics theory could rewrite the textbooks”, published in Physical Review Letters (January 27, 2016), and concerning the experiments made in the Thomas Jefferson National Accelerator, you say:

"For many scientists, the idea that the internal structure of protons might change under certain circumstances can seem absurd, even sacrilegious. To others like myself, evidence of this internal change is highly sought after and would help to explain some inconsistencies in theoretical physics."

http://phys.org/news/2016-01-physics-theory-rewrite-textbooks.html



It’s a good surprise for me to know about the ongoing experiments at Jefferson Lab , because a new structure for the proton is proposed in my book Quantum Ring Theory, published in 2006:

http://www.amazon.com/Quantum-Ring-Theory-Wladimir-Guglinski/dp/0972134948/ref=sr_1_3?s=books&ie=UTF8&qid=1454004785&sr=1-3&keywords=guglinski


The figure ahead shows the structure of the proton, which is the following:

1- A body-ring formed by three quarks: (u,d,u)

2- The body ring is crossed by a flux of gravitons g(+)

3- The rotation (spin) of the body-ring induces a principal field Sp(p)

4 – The rotation of the principal field Sp(p) induces a secondary electric field Sn(p).

5- The rotation of the secondary field Sn(p) induces a Coulomb electric field Sc(p). The Coulomb field does not rotates, it is bound to the frame of the structure of the Universe around the proton.

Image:The_proton_and_its_3_fields.png


The structure of the electron is similar.


Looking at the structure of the proton shown at the previous figure, you may realize why the idea that the internal structure of protons might change under certain circumstances is NOT absurd, since there is an inner asymmetry in its structure, as we see in the figures ahead when the proton is moving:


Image:Different_motions_of_the_proton.png


It is very advantageous to consider seriously this new structure of the proton, because by using this model it’s possible to eliminate several unsolved puzzles of the Fundamental Physics, as shown ahead:



1- The paradox of the Schroedinger Equation

Schroedinger has developed his equation by considering a free electron. Therefore, by considering the atom model of Quantum Mechanics, his equation cannot be applied to the atom, since in the atom the electron is non free. Therefore, the atom model of Quantum Mechanics is incompatible with the Schroedinger Equation.

In order to reconcile the Schroedinger Equation with the atom there is need to find a new model of atom. This new model of atom is proposed in Quantum Ring Theory, and ahead is explained the reason why Schroedinger Equation can be applied to the atom, in spite of he had considered a free electron for development of his theory:


1- The electron always moves with helical trajectory (the zitterbewegung discovered by Schroedinger in the Dirac’s equation of the electron).

2 - The space of the Coulomb field Sc(p) is Euclidian, and it has a density r0 (r0 is the density of the space in the regions far away of any presence of matter). Within an Euclidian space with density r0 the laws of Quantum Mechanics are well applied. So, when a free electron and a free proton are separated by a large distance they interact via electric force of attraction as considered in Quantum Mechanics, and the electron moves with acceleration attracted by the proton.

3- Unlike, the space within the Secondary field Sn(p) is non-Euclidian. The density of the space in a point P within the field Sn(p) follows the equation d = r0/R , where R is the distance between the point P and the center of the proton.

4- When the electron exits the Euclidian space and it enters within the non-Euclidian space due to the field Secondary field Sn(p), where the density of the space is growing along the radial direction toward the proton, the radius of the electron’s zitterbewegung decreases, causing a force between the proton and the electron in contrary direction of the electric force of attraction between them. The two forces have the same intensity, and therefore within the non-Euclidian space the electron moves with constant speed. This is the reason why the Schroedinger Equation can be applied to the atom, because for a non-free electron moving with constant speed within a non-Euclidian space into the atom we can apply the same equation applied for a free electron moving with constant speed within an Euclidian space.


.                                                                                    ***


2- The paradox of the rotation of the even-even nuclei with Z=N at the ground state

Before 2012 the nuclear physicists have believed that the even-even nuclei with equal number Z=N of protons Z and neutrons N have no rotation in the ground state. They also believed that those nuclei have spherical shape, as required from the laws of the Standard Nuclear Physics.

But in 2012 the journal Nature has published the paper How atomic nuclei cluster, describing experiments which detected that even-even nuclei with Z=N have ellipsoidal shape:

http://www.nature.com/nature/journal/v487/n7407/index.html#lt


A nucleus with eliposoidal shape cannot have null electric quadrupole moment. But the experiments show that even-even nuclei with Z=N have zero elec. quad. moment.

So, a paradox has arisen.


In my book Quantum Ring Theory published in 2006 it is predicted correctly that even-even nuclei with Z=N have ellipsoidal shape. Thereby, 6 years before the publication of the paper by the journal Nature I had to explain the paradox: as according to my new nuclear model the even-even nuclei with Z=N have ellipsoidal shace, how can they have zero elec. quad. moment?

The argument proposed by me so that to solve the paradox is explained in the page 137 of my book Quantum Ring Theory.


When the paper How atomic nuclei cluster was published by Nature in 2012, I sent an email to that journal, asking an explanation for the paradox, since due to their ellipsoidal shape those nuclei could not have elec. quad. moment zero.

The nuclear theorist Martin Freer sent me a reply, giving the same explanation proposed in the page 137 of my book QRT. His explanation, like that published in the page 137 of my book, considers that the even-even nuclei with Z=N have rotation in the ground state.

Therefore, according to Martin Freer and the nuclear theorists authors of the paper published by Nature, the even-nuclei with Z=N have rotation in the ground state.


However to consider that those nuclei have rotation in the ground state introduces a new paradox in the Standard Nuclear Physics. Because due to the electric charge of the protons, a magnetic moment is induced by their rotation, and therefore those nuclei cannot have null nuclear magnetic moment. But the experiments have detected that they have zero magnetic moment.


Therefore, by considering the Standard Nuclear Physics is impossible to solve the puzzle.

The puzzle can be solved only by considering the new structure of the proton formed by three fields, as shown in my paper “Aether Structure for unification between gravity and electromagnetism (2015)”:

http://peswiki.com/index.php/Aether_Structure_for_unification_between_gravity_and_electromagnetism_%282015%29


.                                                                                    ***



3- Ra224 pear shape paradox

Even-even nuclei with Z and N different also cannot have rotation at the ground state, because they have null magnetic moment. But an experiment made in the Liverpool University is suggesting that they have rotation, because the nucleus Ra224 has a pear shape impossible to be explained by considering the Standard Nuclear Physics.

Trying to solve the puzzle, Prof. Butler of the Liverpool University suggested that there is a z-axis dividing the nuclei. However, the puzzle remains, because other question cannot be anwered: why are the even-even nuclei divided by a z-axis? After all, there is not any law supported by the Standard Nuclear Model able to explain why they are divided by a z-axis. The existence of the z-azis suggested by Prof. Butler is predicted in the page 133 of my book Quantum Ring Theory, where it is written the following concerning the distribution of protons and neutrons within the atomic nuclei:

The distribution about the z-axis is a nuclear property up to now unknown in Nuclear Physics


Image:Prof_Butler.png


See: 'Pear-Shaped Nucleus Boosts Search for Alternatives to "Standard Model" Physics

http://www.scientificamerican.com/article/pear-shaped-nucleus-boost-search-for-alternatives-to-standard-model-physics/


.                                                                                    ***



4- Puzzle of the Rossi-Effect

As is known, from the laws of the Standard Nuclear Model is impossible the cold fusion occurrence.

Neverthless the cold fusion phenomenon known as “Rossi-Effect” was confirmed by three universities in Europe:


Observation of abundant heat production from a reactor device and of isotopic changes in the fuel

http://amsacta.unibo.it/4084/1/LuganoReportSubmit.pdf


So, the experimental confirmation for cold fusion requires an explanation, and it is impossible to find it by considering the nuclear models of the Standard Nuclear Physics.


The cold fusion occurrence is possible by considering the field model surrounding the proton, because when many protons (and neutrons) compose a nucleus the three fields surrounding that nucleus keep the same configuration like they have in the proton. For instance, the figure ahead shows the nucleus 2He4 surrounded by its three fields:


Image:Nucleus_2He4.png


In the Standard Nuclear Model the cold fusion is impossible because it is considered that the electric field surrounding the nuclei is spherical. So, the energy necessary for a nucleon to enter is the same in any point of the Coulomb barrier, no matter the point where the nucleon tries to cross it.


The figure ahead shows what happens when we consider the new model of field composed by three fields:

a) In normal conditions, the Secondary field Sn(X) of a nucleus X has chaotic rotation. Thereby the Coulomb barrier is spherical (as considered in the Standard Nuclear Physics), and therefore in this normal conditions only hot fusion may occur.

b) But in special conditions, as occurs in the Rossi-Effect, the field Sn(X) of the nucleus X can be aligned toward the z-axis, and so the field Sn(X) stops to gyrate chaotically. The field Sn(X) starts to gyrate about the z-axis, and by this way cold fusion is possible to occur, because a nucleon can cross easily the Coulomb barrier of the nucleus X when the nucleon hits the Coulomb barrier moving along the z-axis.


Image:Changing_of_the_secondary_field.png


The mechanisms for the cold fusion occurrence are shown in details in this paper:

Cold fusion mystery finally deciphered

http://peswiki.com/index.php/Cold_fusion_mystery_finally_deciphered


.                                                                                    ***



5- The proton radius puzzle

Before 2010 the nuclear theorists were sure that proton’s size is the same under any condition, and it has an undeformable radius with length 0,87fm. But in 2010 the journal Nature has published a paper describing a new experiment made according to which the proton size can shrunken, and the new radius obtained was 0,84fm:

Shrunken proton baffles scientists

http://www.nature.com/news/shrunken-proton-baffles-scientists-1.12289


The nuclear theorists think the shrinkage can be due to errors in the measurement, and they hope to eliminate the controversy with the MUSE experiment, to be performed between 2016 and 2017. The MUSE experiment will be made with muons 200 times heavier than the electrons.

According to the new model of proton proposed in Quantum Ring Theory, the radius of the proton depends on the intensity of the flux of gravitions crossing its body-ring.

As the muons to be used in the MUSE experiment are heavier than electrons, during the scattering proton-muon the flux of gravitions crossing the proton’s body-ring will be stronger, and it is expected that its radius can have a very big shrinkage.


In my paper Anomalous Mass of the Neutron, published in my book in 2006 (and also in the Journal of Nuclear Physics in 2011) it is calculated that the proton radius within the heavy nuclei has the order of 0,3fm , and so very shorter than the 0,87fm considered in the Standard Nuclear Model.

Anomalous mass of the neutron

http://www.journal-of-nuclear-physics.com/?p=516


In the case the MUSE experiments detect a proton radius very shorter than 0,87fm (for instance between 0,3fm and 0,6fm) such experimental result will reinforce the new structure of proton proposed in my Quantum Ring Theory. And obviously a proton radius obtained between 0,3fm and 0,6fm will represent the collapse of the Standard Model, requiring a New Physics with new principles missing in the Standard Nuclear Physics.


.                                                                                    ***


Several other unsolved puzzles of the Standard Nuclear Physics can be eliminated when we consider this new structure of the proton. But I hope the examples exhibited here were able to give a good idea on the merit of this new structure proposed for the proton.


Regards

Wladimir Guglinski



 
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Experiments have never detected 3D1 state of deuterium (Score: 1)
by vlad on Friday, February 05, 2016 @ 21:59:03 EST
(User Info | Send a Message) http://www.zpenergy.com

To: Dr. Marco Nardecchia and Dr. Sophie Renner

Experts on physics beyond the Standard Model

University of Cambridge


Dears Dr. Marco Nardecchia and Dr. Sophie Renner


In the interview by Marianne Freiberger entitled LHC glimpses hint of new physics, Dr Nardecchia says about some anomalies unable to be explained by the Standard Model:

"Usually the explanation is that the strong interaction effects aren't properly under control [in the calculations]. But now we have two anomalies, one of them being very clean. This could really mean that there is some new physics here."

https://plus.maths.org/content/lhc-glimpses-hint-new-physics


Aong the last decade several experiments are debunking the Standard Model, and often we read interviews where some academic physicists recognize that new puzzles are requiring the development of a New Physics.

Sometimes the need of a New Physics is not even mentioned. For instance, in the paper The quantum vacuum as the origin of the speed of light the authors propose that the space is filled by particles and antiparticles:

We show that the vacuum permeability and permittivity may originate from the magnetization and the polarization of continuously appearing and disappearing fermion pairs.

http://arxiv.org/abs/1302.6165

Well, a space filled by particles and antiparticles is actually a New Physics, in spite of the authors did not mention it.


However, the need of a New Physics is present along the development of the Standard Model since its birth 80 hears ago, because several unsolved puzzles were pointing out that some fundamental laws are missing in the Standard Model.


For instance, consider the deuteron , formed by proton+neutron.


The proton’s charge distribution is spherical, having electric quadrupole moment zero, Q(p)=0 (confirmed by experiments). The neutron has no charge, and therefore it has electric quadrupole moment is also zero, Q(n)=0 (confirmed by experiments).


Therefore it’s IMPOSSIBLE (by considering the laws of the Standard Nuclear Physics ) for the deuteron to have electric quadrupole moment different of zero. From the laws of the Standard Model the deuteron MUST have zero electric quadrupole moment, Q(D)=0.


But in 1939 an experiment detected that deuteron has Q(D)= 2.73×10^−27cm^2, and therefore at that time the nuclear theorists would have to conclude the following: the way they have adopted for the development of the Standard Nuclear Physics is wrong, and a New Physics is need. And of course at that time some voices started to claim that a New Physics is need.


However, instead of admiting that it is impossible to explain the electric quadrupole moment of the deuteron by considering the model of neutron considered in the Standard Model, the nuclear theorists tried to solve the puzzle by introducing new puzzles.

Let us analyze how the puzzle was solved, by considering that the deuteron in the ground state is a mixture of the states 3S1 and 3D1:


Recently the magnetic moments of the proton, neutron and deuteron have been measured with great precision. It is found that the moment of the deuteron differs from the algebraical sum of the moments of the proton and neutron by an amount much greater than the experimental error. The discovery of the electric quadrupole moment of the deuteron in 1939 already indicated that the ground-state of the deuteron is not of pure 3S character but contains an admixture of 3D. The observed magnetic moment can be explained if the percentage admixture is 4 per cent”.

Nuclear Forces and the Magnetic Moment of the Deuteron

http://www.nature.com/nature/journal/v160/n4075/abs/160794a0.html


The physical meaning of the states 3S1 and 3D1 are shown in the Figure 1:


Image:3S1_and_3D1_states_of_deuteron.png



According to the Standard Model, the deuteron at the ground state is a mixture of the states 3S1 and3D1, as follows:

· 96% of the time the deuteron has no orbital momentum, l=0 ,

· and along 4% of the time it has rotation with momentum l=2


First of all, we have the following fundamental question:

· why a hell the deuteron exits the most stable state 3S1 (with no rotation, l=0), and it starts to rotate, with angular momentum l=2 ??


After all, there is need energy so that to occur the changing from 3S1 to 3D1, and obviously the question is: where the deuteron gests such energy from?


But the solution adopted in the Standard Model is also incompatible with the experiments, as shown in the Figure 2.


Suppose that three experiments A, B, and C, have been made as follows: 100 measurements have been made in each experiment, with the total of 300 measurements:


Image:Figure_2,_experiments_A,_B,_C.png


Figure 2 shows statistical results similar to what we have to expect from the Standard Model, by considering that the deuteron stays 96% of the time at the 3S1 state and 4% of the time at the state 3D1:


In the collection of the experiments A, 94 measurements have to obtain Q(3S1)=0, and 6 measurements have to obtain Q(3D1) = 68,2x10^-27cm^2.


In the collection of the experiments B, 96 measurements have to obtain Q(3S1)=0, and 4 measurements have to obtain Q(3D1) = 68,2x10^-27cm^2.

· In the collection of the experiments C, 98 measurements have to obtain Q(3S1)=0, and 2 measurements have to obtain Q(3D1) = 68,2x10^-27cm^2.


So, in the total, the experiments would have to get:

· 288 measurements with the value Q(3S1) = 0

· and 12 measurements with the value Q(3D1) = 68,2×10^−27cm^2.



Besides, if the deuteron existed in the 3D1 state, when it decays to the state 3S1 it had to emit energy, since it lost the energy of rotation from l=2 to l=0. But emission of photons by the deuteron has never been detected.

Other reason why 3D1 state is impossible to occur is explained ahead.


Let’s apply the equation for the angular orbital moment as function of the magnetic dipole moment for the case of the 3D1 state of deuteron, as shown in the figure of the link ahead:

http://coral.ufsm.br/gef/Moderna/moderna10.pdf


Image:Momento_magnetico.png


The sum of the nuclear magnetic moments of proton+neutron is:

+2,7896 – 1,9103 = +0,8793

The nuclear magnetic moment for the deuteron is +0,857 , measured by experiments.

The difference is +0,8783 – 0,857 = 0,022

But the deuteron stays only 4% of the time at the state 3D1, and therefore in such state it has to produce a negative magnetic moment equal to:

mi = – 0,022 /0,04 mN = - 0,55 miN.


So, from mi = (e/2m).L we have for the deuteron:

mi = [e/2(mP + mN)].L , where:

mP and mN are the masses of proton and neutron

L= (mP + mN).V.R


0,55mN = [1,6x10^-19/2.( mP + mN) ].( mP + mN).V.R



Being 0,87fm the radius of the proton, the radius of the orbit of the proton’s center (and the center of the neutron) up to the center of the deuteron is:


R= 1,95 – 0,87 = 1,08fm

0,55miN = 1,6x10^-19.V.1,08x10^-15/2

But miN = 5,05x10^-27 J/T

0,55 x 5,05x10^-27 = 0,864 x 10^-34 .V

V= 3,215x10^7 m/s


Let us calculate the work of the required force for moving the center of the proton and the center of the neutron along a distance 1,0fm from the center of the deuteron (to move away each one of them).

As the distance between the center of proton to the center of deuteron is 1,08fm, when the center of the proton moves away 1fm from the center of the deuteron (and the center of the neutron also moves away 1fm), then the distance between the two centers of the proton and neutron will be 1,08x2fm + 2fm = 4,16fm, and with such distance between them the strong nuclear force between the proton and neutron will not actuate.


The changing of the centripetal force on the proton (and on the neutron) moving from a radius R= 1,08fm up to a radius R= 2,08fm is:

Fc = mV^2 . int[ 1/R ](1,08 to 2,08)

Fc = 1,67x10^-27 x (32,15x10^6)^2 x [ln2,08 – ln1,08]/10^-15

Fc = 1726 x (0,732 – 0,077) = 1131,3 N


The total displacement “d” traveled by the center of the proton and by the center of the neutron under the action of the force Fc is:


d= 2x10^-15

The work is:

L = Fc.d = 1131,3 x 2x10^-15 = 2263 x 10^-15 J = 22,6x10^-13 J


The deuteron’s binding energy is 2,2MeV = 3,5x10^-13 J , and so it is 6,5 times weaker than the energy produced by the centripetal force Fc to put the proton and neutron with their centers separated by a distance 4,16fm within the deuteron.


Thereby the deuteron at the state 3D1 will not withstands the action of the centripetal forces on the proton and neutron, and the deuteron will break.


This conclusion is reinforced by the fact that deuteron’s 3D1 state was never confirmed by experiments. If the deuteron existed in the 3D1 state, obviously it would be of extreme interest for the nuclear physicists to confirm experimentally its existence, because such confirmation would confirm the theory according to which the deuteron has non null electric quadrupole moment as consequence that it is a mixture of the states 3S1 and 3D1.

If the deuteron existed in the 3D1 state, the experiments would measure the following nuclear properties:

· Magnétic moment

· Spin

· Life-time

· Electric quadrupole moment



In resume, it’s IMPOSSIBLE to explain the electric quadrupole moment and the magnetic moment for the deuteron, by considering the model of neutron adopted in the Standard Nuclear Physics. Because a deuteron formed with the theoretical neutron model proposed in Standard Model cannot have Q= 2,7×10^−31 m^2 and m = +0,857miN.

Actually a deuteron formed by the neutron considered in the Standard Model must have:

· Q= 0

· mi = +0,8793miN


The values Q= 2.7×10^−27 cm^2 and mi =+0,857miN for the deuteron measured by experiments can be obtained theoreticall only by considering a new model of neutron, as proposed in my paper ANOMALOUS MASS OF THE NEUTRON, published in the book Quantum Ring Theory, and also published as a paper in the Andrea Rossi’s blog Journal of Nuclear Physics:


http://www.journal-of-nuclear-physics.com/?p=516


In the paper it is calculated the electric quadrupole moment, getting the value Q= + 2.7×10^–31m^2 , and the magnetic moment getting mi =+0,857miN, without the need of adopting unacceptable nonsenses (similar to the nonsenses adopted in the theory of the deuteron based on the neutron model of the Standard Nuclear Physics).


The nuclear theorists cannot avoid the use of nonsenses in some theories based on the nuclear models of the Standard Model, because as the Standard Model was developed from some mistaken assumptions, it is impossible to avoid puzzles impossible to be solved. And so, any time when a new experiment debunks the Standard Model, the nuclear theorists try desperately to save the theory, by proposing nonsenses. For instance, in 2009 the Physical Review Letters has published the paper “Atomic nucleus of beryllium is three times as large as normal due to halo”.

http://www.uni-mainz.de/eng/13031.php


For the first time, scientists had measured the size of a one-neutron halo with lasers, and the measurement proved that nucleons are not bound within the nuclei by the strong force, because in the 4Be11 the halo-neutron is 7fm far away from the rest of the cluster, and since the strong force actuates in a maximum distance shorter than 3fm, it is obvious that the neutron is not bound via the strong force in the Be11.


Trying to save the Standard Nuclear Theory, the nuclear theorist Dr. Wilfried Nörtershäuse has proposed the following desperate solution:

The riddle as to how the halo neutron can exist at such a great distance from the core nucleus can only be resolved by means of the principles of quantum mechanics: In this model, the neutron must be characterized in terms of a so-called wave function. Because of the low binding energy, the wave function only falls off very slowly with increasing distance from the core. Thus, it is highly likely that the neutron can expand into classically forbidden distances, thereby inducing the expansive 'heiligenschein'.


But beyond the fact that Nörtershäuse’s theory is very strange, because he is proposing a sort of neutron which behaves like a rubber band used by dressmakers, his theory is also unacceptable, because:


1) Suppose that Nörtershäuser’s theory was viable and wise, and the neutron indeed could have the strange property of behaving like a rubber band. However his theory cannot explain other experimental fact: the 4Be11 decay produces the stable isotope 5B11, and there is no way to explain it by considering the Nörtershäuser’s hypothesis.

Indeed, Nörtershäuser’s hypothesis is also unacceptable because of the feature of the decay of the nucleus 4Be11, as explained ahead:


2) He could argue that the halo-neutron is weakly linked to the cluster, and it exits the nucleus after the 13,81 seconds just because of the weak link. However this is no true, because in 97% of decays the 4Be11 transmutes to 5B11, and therefore the neutron does not exits the nucleus. In 4Be11 the neutron decays in a proton and electron, and the proton goes back to the cluster. If the strong nuclear force was responsible for the cohesion of nuclei as the nuclear theorists suppose, the proton could never go back to the cluster, because in a distance of 7fm it cannot interact with the cluster via strong force, and the classical Coulomb repulsion between the cluster and the proton would be so strong that the proton would be expelled from the 4Be11, and 5B11 could not be formed in 97% of the 4Be11 decay.



3) Therefore, even if the bizarre Nörtershäuser’s solution was viable for the explanation of the halo neutron in a distance of 7fm from the rest of the nucleus, however the 5B11 would never be formed from the decay of the 4Be11, according to his solution.


4) And in his paper Nörtershäuser did not propose any explanation for the formation of the isotope 5B11 from the decay of the 4Be11. He only tried to explain how a neutron could be kept in a distance of 7fm. So, Nörtershäuser solution is unacceptable, and therefore it is impossible to explain the 7fm distance of the neutron in the Be11 by considering the current nuclear models based on the Standard Nuclear Physics. The distance of 7fm detected in the experiment suggests that nucleons are not bound in the nuclei via the strong nuclear force, as predicted in Quantum Ring Theory.



.                                                                                       * * *


So, dears Dr. Marco Nardecchia and Dr. Sophie Renner,

concerning the words "But now we have two anomalies, one of them being very clean. This could really mean that there is some new physics here", I hope you may realize that many new experiments made in the last 10 years are showing the need of a New Physics in a way so much strinking, and the most wise decision of the community of physicists must be face the unavoidable:

there is need to review the Standard Model by starting from the beginning,

as for instance from the model of the neutron


Regards

Wladimir Guglinski




 

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