by
Wladimir Guglinski
retired, author of the Quantum Ring Theory
.
.
Abstract
Dr. Wilfried Nörtershäuser of the Helmhotz Center for Heavy Ion Research at the University in Mainz says on the 2009 experiment which had detected a neutron halo in 4Be11 with distance 7fm from the cluster:
“By studing neutron halos, scientists hope to gain further understanding of the forces within the atomic nucleus that bind atoms together, taking into account the fact that the degree of displacement of halo neutrons from the atomic nuclear core is incompatible with the concepts of classical nuclear physics”[ 2 ]
In the case of 4Be11, the halo neutron and the nuclear core are separated by the distance of 7fm, and so such isotope represents the experimental proof that the cohesion of nucleons within the light isotopes cannot be promoted by the strong nuclear force.
Such experimental discovery published in 2009 had been predicted years ago, because according to the new nuclear model proposed in Quantum Ring Theory, published in 2006, the cohesion of the nucleons within the light nuclei is not caused by the strong nuclear force.
Here in this paper the new nuclear model is submitted to a scrutinity so that to verify whether from its structure it’s possible to explain the stability of the light nuclei and to reproduce the nuclear properties as nuclear spins, electric quadrupole moments, and magnetic moments. Nuclear magnetic moments are calculated from two different and independent methods. In the second, named “method of equilibrium between nucleons”, it’s presented the Lagrangian for nuclei with Z < 8. The results obtained from them agree each other, and are corroborated by nuclear spins and electric quadrupole moments suplied by nuclear tables.
In this Part One are presented calculations on magnetic moments for the isotopes of lithium, beryllium, and boron. In the next paper Part Two will be exhibited calculations for carbon, nitrogen, and oxygen. In the paper Part Three the author will exhibit calculations for electric quadrupole moments.
.
.
Joe wrote in April 16th, 2013 at 1:06 AM
Wladimir,
1. You state that the central 2He4 has no rotation but has spin. What do you mean by rotation versus spin?
2. Do all the nucleons (except 2He4) rotate about the z axis?
3. If so, what is the force that carries them?
4. Are the nucleons in continual motion along n(o), or are they immobile?
5. What is the mechanism in QRT that permits a loop of n(o) to act independently of other loops, altering its radius or orientation?
6. What do you mean by “magnetic field of the nucleus”? Is it the sum of all F(M) and F(SI)?
All the best,
Joe
1. Spin is the rotation of a body about its own line center. For instance, the rotation of the Earth about its line center in 24h. Rotation is the orbit about a center. For instance the Earth rotation about the Sun in 365 days.
2. Due to the fact that nucleons are distributed symmetrically about the z-axis which crosses the central 2He4, the tendency to gyrate having the 2He4 as a center is a natural tendency.
Besides, the fluxes n(o) which crosses the central 2He4 contribute for such tendency.
But in the case of the light nuclei which have a very big unbalance of masses, the line center of rotation can suffer a little dislocation. That’s why it’s hard to calculate theoretically the exact their magnetic moment.
The 5B10 is an exception. Look at its structure in the Fig. 39, page 51.
The 3 deuterons are in the side Ana, and they cause a very big mass unbalance. Even the flux n(o) cannot avoid a big displacement of the center of rotation.
The line center of rotation in 5B10 crosses the x-axis in a point situated between the deuteron D-1 and the 2He4.
Then the 2He4 gyrates about the line center, and as it has a charge 2, there is magnetic force on the 2He4 which contributes for the equilibrium of the nucleus.
In the paper PART TWO which I am writing now I calculate the point where the line center crosses the x-axis, so that to calculate the magnetic moment of 5B10 and also its electric quadrupole moment.
3. The forces are calculated in my paper:
Fm = 1,6×10-19.ω.R.µ
where:
ω = angular velocity
R= orbit radius of the nucleon
µ= magnetic moment of the nucleus
4. The flux n(o) gyrates with the nucleus, together with the nucleons. The nucleons have a spin, and are crossed by the flux n(o).
As the flux n(o) is gravitational, one flux n(o) would have to capture the others, and they would not be distributed symmetrically. But each flux n(o) is surrounded by a cloud of repulsive gravity particles G, which avoid the attraction of the fluxes n(o) among them.
5. The fluxes n(o) are created by each deuteron of the central 2He4, see Figure 1. That creation of the fluxes rules their behavior.
6. The magnetic moment of a nucleus is caused by a magnetic field
regards
wlad
Dear James Bowery:
Yes, we are on our way to certify the domestic E-Cat .
Warm Regards,
A.R.
Wladimir,
1. You state that the central 2He4 has no rotation but has spin. What do you mean by rotation versus spin?
2. Do all the nucleons (except 2He4) rotate about the z axis?
3. If so, what is the force that carries them?
4. Are the nucleons in continual motion along n(o), or are they immobile?
5. What is the mechanism in QRT that permits a loop of n(o) to act independently of other loops, altering its radius or orientation?
6. What do you mean by “magnetic field of the nucleus”? Is it the sum of all F(M) and F(SI)?
All the best,
Joe
New experiment(April 2013) corroborates proton’s syze calculated in Quantum Ring Theory
From: wladimirguglinski@hotmail.com
To: johna_6@yahoo.com
CC: epja@itkp.uni-bonn.de; helayel@cbpf.br; jyeston@aaas.org; prc@aps.org; apr-edoffice@aip.org; nature@nature.com; cjp@fzu.cz; ver@cisp-publishing.com; pnj@bauuinstitute.com
Subject: Proton Size Smaller Than Physicists Thought, Puzzling New Measurements Suggest
Date: Mon, 15 Apr 2013 23:41:31 -0300
To: Dr. John Arrington
Argonne National Laboratory
Subject: Proton Size Smaller Than Physicists Thought, Puzzling New Measurements Suggest
Dear Dr. John
In my paper ANOMALOUS MASS OF THE NEUTRON it is calculated that the proton’s radius is Rp= 0,275fm (page 100 of my book Quantum Ring Theory, published in 2006).
In 2002 the paper been rejected by the Czechoslovak Journal of Physics, with the following reports (by the editor-in-chief and the Referee):
==================================
Prague, 16 May 2002
Ref: CJP-5263
Paper: Anomalous mass of the neutron
Author(s): Guglinski W.
Dear Dr. Guglinski,
we are sorry to inform you that your above paper has not been recommended for publication (see enclosed Referee’s Comments).
With best regards,
P Pavlo
Editor in Chief
==================================
==================================
Referee Report on MS CJP – 5263
Anomalous Mass of the Neutron
by W. Guglinski
The paper deals with an artificial construction of the neutron considered as the bound state of the proton and electron.
There are plenty of well established facts and considerations which strongly contradict claims and deductions of the paper:
A) proton radius is of about 0,8fm
B) deuteron radius is of about 2,4fm
C) magnetic dipole moment and quadrupole electric moment of the deuteron are well explained in modern nucleon-nucleon force models by the presence of the tensor component which is well confirmed alson from scattering data
I recommend not to publish paper in Czechoslovak Journal of Physics
====================================
My paper ANOMALOUS MASS OF THE NEUTRON had been also published in Andrea Rossi’s Journal of Nuclear Physics, in 2012:
http://www.journal-of-nuclear-physics.com/?p=516#more-516
Now the experiments are suggesting that the proton’s size calculated in my paper can be correct:
Proton Size Smaller Than Physicists Thought, Puzzling New Measurements Suggest
http://www.huffingtonpost.com/2013/04/14/proton-size-smaller-physicists-new-measurements_n_3080196.html?utm_hp_ref=science
In 3th October 2012 you sent me an email where you wrote to me, concerning the new experimental findings published in 2012 in the branch of Nuclear Physics:
“I am not aware of any deficiencies in the current models, and in particular, not in the context of our recent measurement. That does not mean that there aren’t any deficiencies, but I’m not going to believe that these common and well-tested models are wrong simply because you say so and provide a hand-waving argument.”
But now, in the link above, you say:
“Most exciting of all, the discrepancy could reveal some new physics not explained by the dominant physics theory, the Standar Model. Perhaps there is something unknown about how muons and electrons interact with other particles”
So, please tell to me, dear Dr. Arrington:
did you change your mind ?
Regards
Wladimir Guglinski
15th April 2013
A batch of chicken eggs in my incubator are going to hatch in a week. The propane (we’re in a farm house in Iowa) is running out due to the cold spring so we have to keep the house temperature down near 60F. The hatchlings need a brooder near 90F for until their feathers come in. I sure could use an E-Cat!
orsobubu wrote in April 15th, 2013 at 3:45 AM
1- Wladimir, is yesterday proton size’s news of any relevance regarding your theory?
Yes, dear orsobubu
The paper ANOMALOUS MASS OF THE NEUTRON, where it is calculated the radius of the proton, had been published by JNP, in 2012:
http://www.journal-of-nuclear-physics.com/?p=516#more-516
The calculation on the radius of the proton is in the item Proton’s radius
I will try to find the report of the referee who rejected the publication of my paper (more or less in 2000), and the name of the reputable journal.
regards
wlad
Joe wrote in April 14th, 2013 at 3:23 PM
Wladimir,
If the neutron is kept within the nucleus by the spin-interaction force F(SI) acting as the centripetal force F(C) on it by nucleons elsewhere in n(o), why can it not also be kept within the nucleus by the F(SI) of the central 2He4 alone?
All the best,
Joe
because there is no magnetic force on the neutron (since it has no charge). Then the neutron is expelled by the centripetal force on it, because there is no magnetic force so that to cancell the action of the centripetal force.
regards
wlad
Joe wrote in April 15th, 2013 at 2:46 AM
Wladimir,
1. You write that “the neutron n-3 is captured by D-2″ (page 69) and “D-2 is pulled by n-3″ (page 70) and “the neutron n-2 goes to the outer side of Ana, and it pulls D-2 and n-3″ (page 71)in the case of 4Be11. Is the intrinsic magnetic moment F(SI) the cause of these actions?
2. Why can not the F(M) of 2He4, which is caused by its rotation with charge +2, interact with the intrinsic magnetic moment F(SI) of a neutron and act essentially as its centripetal force F(C)?
3. There seems to be a contradiction between the following statements:
statement 1: “Then there is magnetic force attraction between the magnetic moment of the halo neutron and the magnetic moment of the cluster. But the centripetal force is stronger, and the radius orbit continues increasing by quantized steps.” (page 72)
statement 2: “There is no magnetic interaction when the halo neutron is close to the cluster, within the structure of 4Be11. There is magnetic attraction between the halo neutron and the cluster when they are separated by a distance of 7fm.” (page 73)
Besides, why would the first sentence of statement 2 be true at all?
4. Why must n-2 of 4Be11 necessarily decay (to a proton, releasing an electron) when it moves further out (at 7fm) from the cluster of nucleons?
All the best,
Joe
Dear Joe
1-
a) D-2 is pulled by n-3 (page 70)
Explanation:
The centripetal force on n-3 is big, because it orbit radius is big. As n-3 is tied to D-2 by the spin-interaction, D-2 is pulled togheter with n-3 by the centripetal force
b) the neutron n-2 goes to the outer side of Ana, and it pulls D-2 and n-3″ (page 71)
Explanation:
When 4Be10 captures the neutron n-3, in the newborn 4Be11 the neutron n-3 has to take a place according to Pauli’s Principle and the Least Action Principle
2-
A charge moving in a magnetic field is submitted to a force. Its a law:
http://webphysics.davidson.edu/physlet_resources/bu_semester2/c12_force.html
a) The 2He4 is central.
It has not rotation. It has not an orbit in the magnetic field of the nucleus.
The central 2He4 has spin. But not a motion within the magnetic field.
b) The central 2He4 has not magnetic field (it’s null). So it cannot interact with a neutron.
3-
I will use an anology with the Earth, so that you may understand it easily:
a) If you go to the center of the Earth, in its center you will not experience any gravitational force on your body.
b) If you start to move away of the center of the Earth, the action of gravity increases with the distance between you and the center.
c) The force of gravity is maximium in the surface of the planet
d) If you leave away the planet, the force of gravity decreases with the distance between you and the center of the planet.
4-
I think there are several statistical variable within the light nuclei, which influence the neutron decays in a proton.
Let’s analyse the causes of the neutron’s decay:
B) A free neutron:
As you know, in my theory the neutron is formed by proton+electron.
The electron moves with a linear Newtonian trajectory (without helical trajectory). Its spin is zero.
The electron is kept moving about the proton thanks to the Coulomb attraction, helped by the magnetic force on the electron (as the electron has a charge, and the neutron has a magnetic field, then the electron gyrating into the magnetic field induces a magnetic force, which helps the Coulomb force).
But the electron’s speed about the proton decreases by quantum steps along the time. A boson W is emitted, and the electron’s speed decreases. A second boson W is emitted, and the speed decreass again.
After 15 minutes the electron’s speed is low sufficiently for the electron to get back its helical trajectory and its spin. In this process an antineutrino is emitted, and the neutron decays, with the electron leaving out the proton.
B) A neutron within light nuclei
There are two situations:
B-1) In the place occupied by the neutron its magnetic moment is in the contrary direction of the magnetic field of the nucleus. In this case the magnetic force on the electron is AGAINST the Coulomb force with keep the proton and the electron together. So, instead of to help the stability of the neutron, unlike the magnetic field of the nucleus causes the decay of the neutron.
B-2) In the place occupied by the neutron its magnetic moment has the same direction of the magnetic field of the nucleus. So the two magnetic fields increase the force of coesion between the proton and the electron, and such neutron gets stability within the nucleus.
B-3) As there is internal oscillations in the internal fluxes which cause the magnetic moment of the nuclei, of course those oscillations can influence the neutron’s decay. Here the process is statistical, and depends on several internal conditions.
Regards
wlad
orsobubu wrote in April 15th, 2013 at 3:45 AM
1- Wladimir, is yesterday proton size’s news of any relevance regarding your theory?
2- “Most exciting of all, the discrepancy could reveal some new physics not explained by the dominant physics theory, the Standard Model. Perhaps there is something unknown about how muons and electrons interact with other particles, said John Arrington, a physicist at Argonne National Laboratory in Illinois”
Dear orsobubu,
1- The experiment of 2009 had measured the proton radius as being Rp = 0,84fm.
The radius of proton is calculated in the page 100 of my book Quantum Ring Theory, and its value is Rp = 0,275fm.
Some years ago a referre had rejected my paper ANOMALOUS MASS OF THE NEUTRON by claiming that the proton’s radius calculated in my paper is very lower than the detected in the experiments.
2- What John Arrington said is a surprise to me, because in 2012 he had replied to me by email that he does not see any problem with the standard nuclear physics (a response to my email that the new experiments published in 2012 in the field of nuclear physics cannot be explained by the current theories, and so a New Nuclear Physics is required).
Look at to Arrington’s words, in his email to me of 3 Oct 2012:
“I am not aware of any deficiencies in the current models, and in particular, not in the context of our recent measurement. That does not mean that there aren’t any deficiencies, but I’m not going to believe that these common and well-tested models are wrong simply because you say so and provide a hand-waving argument.
regards
wlad
To the Readers:
For commercial issues please contact
info@leonardocorp1996.com
We are spamming all the comments requesting commercial information.
Warm Regards,
A.R.
Wladimir, is yesterday proton size’s news of any relevance regarding your theory?
“Most exciting of all, the discrepancy could reveal some new physics not explained by the dominant physics theory, the Standard Model. Perhaps there is something unknown about how muons and electrons interact with other particles, said John Arrington, a physicist at Argonne National Laboratory in Illinois”
http://www.huffingtonpost.com/2013/04/14/proton-size-smaller-physicists-new-measurements_n_3080196.html?utm_hp_ref=science
Wladimir,
1. You write that “the neutron n-3 is captured by D-2” (page 69) and “D-2 is pulled by n-3” (page 70) and “the neutron n-2 goes to the outer side of Ana, and it pulls D-2 and n-3” (page 71)in the case of 4Be11. Is the intrinsic magnetic moment F(SI) the cause of these actions?
2. Why can not the F(M) of 2He4, which is caused by its rotation with charge +2, interact with the intrinsic magnetic moment F(SI) of a neutron and act essentially as its centripetal force F(C)?
3. There seems to be a contradiction between the following statements:
statement 1: “Then there is magnetic force attraction between the magnetic moment of the halo neutron and the magnetic moment of the cluster. But the centripetal force is stronger, and the radius orbit continues increasing by quantized steps.” (page 72)
statement 2: “There is no magnetic interaction when the halo neutron is close to the cluster, within the structure of 4Be11. There is magnetic attraction between the halo neutron and the cluster when they are separated by a distance of 7fm.” (page 73)
Besides, why would the first sentence of statement 2 be true at all?
4. Why must n-2 of 4Be11 necessarily decay (to a proton, releasing an electron) when it moves further out (at 7fm) from the cluster of nucleons?
All the best,
Joe
Wladimir, Thanks for your reply and I kmow what you are saying is correct. Maybe I did not include sufficient information. This subject is difficult to relay. I agree the radius of the helical trajectory diminishes with the growth of the velocity of the particle. In the hydrogen atom the electron moves with helical trajectory because its radius orbit about the proton is very large and therefore not confined in a very short space You are correct and therefore I agree. I shall send more information to explain more and I appreciate your comments. My concern with this subject is the structuring of aether and the aether structure that results from it and the consequentual structure that forms atomic substance. Anyway I am putting some material together and would appreciate your comments on it. All the best Eric Ashworth.
Report from Journal of Physics G on my paper Stability of light nuclei
===================================
To: wladimirguglinski@hotmail.com
Subject: Final decision on your article from J. Phys. G: Nucl. Part. Phys. – G/466076/PAP/111441
From: jphysg@iop.org
Date: Mon, 11 Mar 2013 14:29:16 +0000
Ref: G/466076/PAP/111441
11 March 2013
Dear Dr Guglinski
TITLE: On the stability, magnetic moments, nuclear spins, and electric quadrupole moments of light nuclei with Z < 9 – Part One
AUTHORS: Dr Wladimir Guglinski
Thank you for your submission to Journal of Physics G: Nuclear and Particle Physics. However, we do not publish this type of article in any of our journals and so we are unable to consider your article further.
Thank you for considering Journal of Physics G: Nuclear and Particle Physics.
Yours sincerely
Tom Slader
Publishing Administrator
Journal of Physics G: Nuclear and Particle Physics
Publishing Team
Tom Slader – Publishing Administrator
Ben Sheard – Publishing Editor
Rachel Lawless – Publisher
Jacky Mucklow – Production Editor
===================================
My reply to Journal of Physics G:
From: wladimirguglinski@hotmail.com
To: jphysg@iop.org
Subject: RE: Final decision on your article from J. Phys. G: Nucl. Part. Phys. – G/466076/PAP/111441
Date: Mon, 11 Mar 2013 19:52:30 -0300
Dear Tom Slader
Sorry for my mistake.
I had wrongly supposed that Phys. G: Nucle Part. Phys. should be a serious scientific journal, devoted to publication of papers written according to the scientific method.
My new nuclear model is able to explain the nuclear properties of the light isotopes, and the model is corroborated by the last experiments published between 2009 and 2012.
Then I conclude that your journal publishes only those fantasy nuclear models, incompatible with the nuclear properties of the light nuclei, and denied by the data supplied by nuclear tables.
Regards
WLADIMIR GUGLINSKI
Wladimir,
If the neutron is kept within the nucleus by the spin-interaction force F(SI) acting as the centripetal force F(C) on it by nucleons elsewhere in n(o), why can it not also be kept within the nucleus by the F(SI) of the central 2He4 alone?
All the best,
Joe
Dear Wladimir:
The certification protocols for households do not work that way. We are following the procedure suggested by the Certificators. The procedure suggested by you does not resolve the core of the problem, which is the domestic use, made by unskilled persons in normal houses, without any particular precaution. We already have obtained the certification for the industrial plants, the safety of the E-Cats has already been certified in safe environments, where professional structures and persons are involved.
Warm Regards,
A.R.
Andrea Rossi wrote in April 12th, 2013 at 3:54 PM
Dear Claud:
No, we will put the domestic E-Cat in the market, in due time: certification and recognition of the IP are critical issues.
Warm Regards,
A.R.
Dear Andrea,
I think it is hard to get the certification because you are trying to get it for the E-Cat itself.
But there is an alternative way, as follows: to get a certification for the safe operation of the E-Cat, and not for the E-Cat itself.
So, you have to elaborate a safe procedure for its operation.
For instance:
PROCEDURE FOR SAFE OPERATION OF DOMESTIC E-CATS:
1- The domestic E-Cats must be installed by keeping a minimum distance of 100 metters from any afflux of people.
2- The E-Cats must be installed 1 meter under the ground, within a box with the dimensions 0,5m x 0,5m x 0,5m , which walls must be covered with lead with thickness 5mm.
3- The place of instalation must be isolated by a 2 metters height wall, the door must be closed with a lock, and a plate of warning: KEEP OUT
Of course the cost of instalation will be very higher then the cost of the own E-Cat. Probably between 5 to 10 times higher.
However many people will be interested to buy your domestic E-Cat, and to install it according the SAFE PROCEDURE, because all we are eager of seeing your domestic E-Cat working.
After one or two years of operation according to such safe procedure, you will prove that the domestic E-Cat does not need such a high cost procedure.
And then you will able to get the certification for the own E-Cat, independent of the procedure of instalation.
regards
wlad
Dear Joe,
Look at the structure of 4Be8 in the Figure 14.
We have:
1- It has magnetic moment zero, and therefore there is no magnetic forces on the two deuterons.
2- But the two deuterons are captured along the z-axis, and therefore the radius of their orbit is zero, and so there is no centripetal force on them
3- So, due to the spin interaction between the two deuterons, they enter within the central 2He4, and the 4Be8 decays in two alpha particles (two 2He4 daugther isotopes).
5- The half-life is 6×10^-17 seconds
Now consider the 6C8.
Its structure is similar to the 4Be8 (but instead of two deuterons, int the place of each deuteron there is a pair proton-proton).
We have:
1- It has magnetic moment zero, and therefore there is no magnetic force on the 4 protons
2- They are captured along the z-axis, and therefore there is no centripetal force on them
3- But there is Coulomb repulsion between the two protons each form a pair proton-proton. Therefore one of them takes the direction going to the central 2He4, and the other is expelled from the nucleus
The same happens to the other pair proton-proton.
Therefore 6C8 emits two protons, and two protons goes to the central 2He4.
4- Like in the case of 4Be8, the daugther isotopes are two 2He4
5- The half-life of 6C8 is 2×10^-21 seconds, shorter than the half-life of 4Be8, because the two protons are expelled faster.
Just because they have similar structures (with magnetic moment zero, nuclear spin i=0, Z= pair, A=pair), their decay is similar, and their half-life is very short.
Up to the present days, the reason why 4Be8 decays is not explained from the concepts of the standard nuclear physics.
regards
wlad
Eric Ashworth wrote in April 13th, 2013 at 11:39 PM
Wladimir, with regards Joes question April 11th 2013. Perhaps, have you considered two types of radial dimensions?. Centripetal that is of a helical trajectory and the other being a spin interaction of four forces, this being linear. One is with regards to the interior of a structure, this being radial within of a helical trajectory i.e curvature force energy. The other being radial a linear radial, this being an exterior force of a structure and thereby a fast vibration.
Dear Eric,
No.
Within the nuclei the nucleons do not move with helical trajectory, because they are confined in very short space.
Within the nuclei they move with classical newtonian trajectory.
The free particles move with helical trajectory, as the electron, the proton, the photon, etc.
But the radius of the helical trajectory diminishes with the growth of the velocity of the particle.
When the particle has a velocity near to the speed of light, and it is confined in a short space (as occurs with the electron captured by a proton, and they form a neutron), the electron loses its helical trajectory.
Within the structure of the neutron the electron moves with the classical newtonian trajectory.
In the hydrogen atom the electron moves with helical trajectory, because its radius orbit about the proton is very large, and therefore it is not confined in a very short space.
Regards
wlad
Wladimir, with regards Joes question April 11th 2013. Perhaps, have you considered two types of radial dimensions?. Centripetal that is of a helical trajectory and the other being a spin interaction of four forces, this being linear. One is with regards to the interior of a structure, this being radial within of a helical trajectory i.e curvature force energy. The other being radial a linear radial, this being an exterior force of a structure and thereby a fast vibration. For aether to form a structurte it has to conform to a geometric pattern, consequently a pattern will interact with a pattern but the radial will conform to the pattern. When a pattern is complete it has an identy and it displays linear radial and it is this that provides it with an identity i.e. a stand alone unit of aether. It is a linear radial of a primary dimension that forms structure and that then forms interior structure of curvature force energy. This subject is difficult to transmit because of a language problem. Your theory is correct but this is why you have a problem. I only accessed your material a few days ago but I shall be putting forward some thoughts with regards to gravity/space which controls aether and which I feel is the crux of the matter. I know you mentioned that this subject requires group thought and I could not agree more so with you. All the best Eric Ashworth.
Dear Joe,
1- did you see the explanation on how the halo neutron in 4Be11 can be kept in a distance of 7fm ?
It is shown in the end of the article (see page 71: Interpretation on the 2009 experiment).
What did you guess about the explanation?
2- did you understand that it’s IMPOSSIBLE to explain it from the current classic principles of Nuclear Physics?
The neutron leaves out the cluster, going to move in a orbit of about 7fm.
From the principles of standard nuclear physics, the neutron would have to leave out the nucleus, because the strong force cannot keep it in a distance of 7fm.
But the neutron does NOT leaves out the 4Be11. And what is worst: the neutron decays in a proton.
And because there is a big Coulomb repulsion between the proton and the cluster, then such proton in a distance of 7fm from the cluster would have to be expelled, and 4Be11 would have to be transmuted to 3Li7 in 100% of the decays
This is just what we had to expect from the standard Nuclear Physics.
However, in 97% of the 4Be11 decays, the proton goes back to the cluster, and the 4Be11 transmutes to 5B11.
It’s IMPOSSIBLE to explain how the proton goes back to the cluster, by considering the Standard Nuclear Theory.
But my theory explains it very well, by considering the magnetic force on the proton. As its mass is half of the mass of a deuteron, the centripetal force on it is small compared with the magnetic force (because the proton has the same charge of the deuteron).
That’s why the proton goes back to the cluster of the 4Be11, and it transmutes to 5B11.
regards
wlad
Joe wrote in April 13th, 2013 at 3:18 PM
Wladimir,
1. A centripetal force is by definition a centre-seeking force. In cases involving charged nucleons (protons, deuterons), the magnetic force F(M) obviously acts as a centripetal force. But in the case of neutral nucleons (neutrons), there is NO apparent force that is responsible for keeping them within the nucleus. This is a problem.
2. If you are now claiming that F(SI) is actually a magnetic force caused by the spin of a nucleon instead of the standard interaction between strong nuclear force and angular momentum, then how does a neutral nucleon (neutron) gain F(SI) since, as you claim on page 7, “it is no able to yield a magnetic force within the magnetic field of a nucleus, because it has no charge.”
All the best,
Joe
1. Joe, obviously you did not understand the paper. The neutron is always kept within light nuclei thanks to its spin interaction with deuterons (or protons), because the flux n(o) alone cannot keep neutrons within the nuclei. That’s why there no exist 2He5.
In order to understand how neutrons are kept within the light nuclei, for instance look at the Figure 13 of the paper the spin force Fsi (green) which connects the neutron to the deuteron, in the structure of the 3Li7.
2. The magnetic force on a particle with circular motion within a magnetic field is caused by its charge. As the neutron has no charge, there is no magnetic force on it due to its rotation within the magnetic fields of the nuclei.
But as the neutron has magnetic moment, its spin gets spin interaction with the deuteron (or with a proton, or even with another neutron).
Regards
wlad
Dear Luca Salvarani:
1- I absolutely have no idea of the date of the publication, I am no more in contact with the Professors. They told me after the tests that the publication could be made by April, but ” could be” does not mean “will be”.
2- I have to be calm, because it is necessary to work well. About the results, I will know them the same day you will: we both will read the report as soon as it will be published.
3- Let’s hope that they will make us resurrect, sort of Lazarus !
Thank you for your enthusiasm,
Warm Regards,
A.R.
Dear Giuliano Bettini:
To celebrate before the results is dangerous, under a scaramantic point of view.
Let’s wait and stay silent, allowing the Professors complete their indipendent work . For now better drink water.
Warm Regards,
A.R.
Wladimir,
1. A centripetal force is by definition a centre-seeking force. In cases involving charged nucleons (protons, deuterons), the magnetic force F(M) obviously acts as a centripetal force. But in the case of neutral nucleons (neutrons), there is NO apparent force that is responsible for keeping them within the nucleus. This is a problem.
2. If you are now claiming that F(SI) is actually a magnetic force caused by the spin of a nucleon instead of the standard interaction between strong nuclear force and angular momentum, then how does a neutral nucleon (neutron) gain F(SI) since, as you claim on page 7, “it is no able to yield a magnetic force within the magnetic field of a nucleus, because it has no charge.”
All the best,
Joe
Dear Andrea,
some of my “Facebook friends” are buying strange potions to celebrate the release of the Report.
There is talk of Beer, Champagne, Asti Spumante, spanish “Moscatel”. I do not agree. To celebrate, Fermi opened a bottle of Chianti. In this case I would have suggested Lambrusco, but perhaps with Lambrusco they should eat something.
If you can tell us, with what you celebrate? Champagne, Whisky, Lambrusco, fresh water, nothing?
(I shall share with them).
Thanks,
Giuliano.
Dear Andrea
1) Sorry but I’m really too eager to read the third party report and discover your new generation hot cat. Can we still expect its publication in April or a bit later? Thank you and forgive my impatience…
2) What are your feelings for the report release, and for the plant delivery? You seem very calm and confident… are you?
PS Please can you tell the professors involved in the third party report, that they’re literally killing us?
Wladimir Guglinski, may I just say my feelings are so strong for your attempt to confront the establishment.
Science in general has fallen to a level of absurdity.
Good luck and best wishes in your battle for Truth.
julian_becker,
this rocket could be propelled by nuclear fusion:
http://www.huffingtonpost.com/2013/04/11/nuclear-fusion-rocket-mars-deep-space-travel_n_3059992.html?utm_hp_ref=science
Report by the editor of Physical Review C, rejecting my paper Stability of Light Nuclei
=========================================
Date: Wed, 27 Feb 2013 13:48:30 -0500
From: prc@aps.org
To: wladimirguglinski@hotmail.com
Subject: Your_manuscript CB10370 Guglinski
Re: CB10370
Stability, magnetic moments, nuclear spins, and electric quadrupole moments of light nuclei with Z<9: Part one
by Wladimir Guglinski
Dear Dr. Guglinski,
We have examined your manuscript which you submitted to Physical Review C. We regret to inform you that your manuscript is not considered suitable for publication in the Physical Review. The manuscript does not present work at a level comparable to present-day research in nuclear physics.
As the manuscript is written, the work comes across as a conjecture rather than as a solid piece of rigorous scientific work. The manuscript does not give one confidence that the work satisfies the high standards that one expects of current research in nuclear physics. Meeting such standards is a requirement before a manuscript can be considered for publication in any of the Physical Review journals. Your manuscript fails to pass this initial test. Therefore, we are not considering your manuscript further.
Yours sincerely,
Benjamin F. Gibson
Editor
Physical Review C
Email: prc@ridge.aps.org
http://prc.aps.org/
==========================================
My reply to the editor of Physical Review C :
From: wladimirguglinski@hotmail.com
To: prc@aps.org
CC: johna_6@yahoo.com; m.freer@bham.ac.uk; josehelayel@gmail.com; noerters@uni-mainz.de; j.dunning-davies@hull.ac.uk
Subject: RE: Your_manuscript CB10370 Guglinski
Date: Fri, 1 Mar 2013 20:46:46 -0300
Dear Dr. Benjamin Gibson
Editor, Physical Review C
In your report you state that my “work come across a conjecture”.
You are wrong. My work had been a conjecture in 2006, when my book Quantum Ring Theory had been published. But today my work is not conjecture, because what you call “conjectures” have been confirmed by experiments, published after the publication of my book in 2006. Therefore, today what you consider as “conjecture” is actually some correct theoretical prediction, confirmed by the experiments, as follows:
1- The light nuclei with Z=N=pair have non spherical shape, as predicted in my work, and confirmed by experiments, although along more than 80 years those works which “satisfy the high standards that one expects of current research in nuclear physics” had predicted wrongly that light nuclei with Z=N=pair have spherical shape.
2- The aggregation of nucleons within the light nuclei is not promoted by the strong nuclear force, as correctly predicted in my work, and now confirmed by the experiment published in 2009, although along more than 80 years those works which “satisfy the high standards that one expects of current research in nuclear physics” had predicted wrongly that light nuclei are linked thanks to interactions by the strong nuclear force.
Therefore you are wrong when you consider that my “work comes across a conjecture”.
But suppose you should be right (but you are not), and suppose the correct predictions of my work should be conjectures, as you claim. However the light nuclei exhibit some nuclear properties which are IMPOSSIBLE to be conciliated with the classical principles of the Nuclear Physics that rule the working of any current nuclear model considered by you as a “solid piece of rigorous scientific work”. For instance:
1- The excited 6C12 has null magnetic moment =0 and nuclear spin i=2. Well, it’s IMPOSSIBLE to conciliate those two nuclear properties by considering any conjecture supported by any current nuclear models proposed as a solid piece of rigorous scientific work.
2- The isotope 4Be7 has null electric quadrupole moment, but it A=7 (odd), it has non null magnetic moment = -1,399, and it has nuclear spin i=3/2. Again, it’s IMPOSSIBLE to conciliate those nuclear properties by considering the classical principles of the Standard Nuclear Physics. If you know any conjecture suported by a nuclear model proposed as a solid piece of rigorous scientific work, please tell me, because I would delighted hear that.
3- The isotpes 6C9, 6C13, 6C15, and 6C17 have null electric quadrupole moment, but they have A=odd, they have non null magnetic moment, and they have non null nuclear spin. Again, it’s IMPOSSIBLE to conciliate those nuclear properties by considering ANY CONJECTURE supported by a solid piece of rigorous scientific work, capable to explain how those four carbon isotopes can have null electric quandrupole moment, since they cannot have a spherical distribution of charge, and they have not =0 and i=0.
Therefore, Dr.Gibson, what you consider as a “solid piece of rigorous scientific work” is actually a fallacy, because it’s IMPOSSIBLE to propose ANY CONJECTURE supported by the principles of the Standard Nuclear Physics with the aim to explain many properties of the light nuclei.
If you dont know ANY CONJECTURE capable to explain the anomalies of the light nuclei, you have not the right of rejecting any conjecture proposed in a work developed from a coherent method of reasoning, mainly in the case of the conjecture to be confirmed by experiments, as occurs with some of the conjectures of my work.
Eisntein also had proposed many conjectures in his papers of 1905 and 1916. He did not propose them because he loved to propose conjectures. He proposed conjectures because he had NO CHOICE, since his conjectues had been the unique possible solution capable to solve the paradox raised up by the Michelson-Morley experiment. If should be possible to solve the paradox by considering a conjecture based on the Newtonian principles prevailing at that time, then of course Einstein would not had proposed his conjectures.
Today Einstein’s conjectures are not conjectures, because they have been confirmed by experiments along years. But if Einstein had written his two papers today, they both would be rejected by any editor of any reputable journal of Physics, with allegation that his “work comes across a conjecture rather than as a solid piece of rigorous scientific method”. And then his theory would never be tested by experiments, and his conjectures would never become correct predictions.
Therefore, Einstein’s conjectures have been tested because his papers had been published.
We live today a similar situation. We have no choice, we cannot avoid new conjectures, because there is NO WAY to propose conjectures so that to explain the nuclear propeties of the light nuclei, by keeping the classical principles of the Standard Nuclear Physics.
My nuclear model is the unique work in which IT IS POSSIBLE to allow conjectures so that to explain the nuclear properties of the light nuclei, because my new nuclear model does not work through the principles of the Standard Nuclear Physics. And some of the conjectures had been confirmed by experiments.
Your decision, rejecting the publication of my work, represents a lamentable attempt of trying to stop the advancement of science.
Regards
WLADIMIR GUGLINSKI
Joe wrote in April 12th, 2013 at 3:02 PM
Wladimir,
1. The question that I was asking concerned the NATURE of the centripetal force in QRT. For example, a roller coaster track could be the source of a centripetal force. Planets and stars have weak gravity as the source of their centripetal force. What specifically is the nature of the centripetal force in QRT?
2. The spin-orbit interaction between nucleons involves angular momentum and the strong nuclear force. But the spin-interaction (SI) force between the nucleons in QRT can NOT involve the strong nuclear force because the distance between the nucleons is too great; they are not in close proximity to each other but spread out along the strong gravitational flux line n(o). Therefore, the SI force can NOT be a real force.
All the best,
Joe
1. Centripetal force is the tendency of a body to continue in a rectilinear motion due to its inertia. Such tendency is given by F= mV^2/R.
2. Strong spin-interaction in the light nuclei is not the known strong nuclear force. Strong spin-interaction is a magnetic force, caused by the spin. As proposed in my paper, it seems that the alignment of spins in the same direction (parallel spins) in the same side of a nucleus (for instance, in the side Douglas) causes a growth in the intensity of the force interaction. That’s why I call it strong spin-interaction. Unlike, there are nucleons which are weakly connected by the spin-interaction. In this case it is not a strong spin-interaction.
I could not suppose that strong spin-interaction should be the strong nuclear force considered in classic nuclear physics, since in my nuclear model the nuclei agglutination is not caused by the strong nuclear force.
regards
WLAD
Dear Claud:
No, we will put the domestic E-Cat in the market, in due time: certification and recognition of the IP are critical issues.
Warm Regards,
A.R.
Wladimir,
1. The question that I was asking concerned the NATURE of the centripetal force in QRT. For example, a roller coaster track could be the source of a centripetal force. Planets and stars have weak gravity as the source of their centripetal force. What specifically is the nature of the centripetal force in QRT?
2. The spin-orbit interaction between nucleons involves angular momentum and the strong nuclear force. But the spin-interaction (SI) force between the nucleons in QRT can NOT involve the strong nuclear force because the distance between the nucleons is too great; they are not in close proximity to each other but spread out along the strong gravitational flux line n(o). Therefore, the SI force can NOT be a real force.
All the best,
Joe
Dear Andrea, after months of reflections I’m concluding that domestic e-cat will be probably never available for the market.
While I think that your activity is going well and that big cats will be properly sold to big customers, I think that you’re mostly aiming at that segment and, besides the actual difficulties with the certificators, the power of the energy grids owners and their powerful lobbies will discourage any effort in the direction of the personal equipment, that means a tremendous threat to their business. And sooner or later you’ll be forced (or lured) to give up.
Am I going wrong? And in what?
My best wishes
Claud
Dear Brian:
I hope too that this finds you well.
Yes, it will be delivered on the 30st of April. We will respect the scheduled term of delivery.
Warm Regards,
A.R.
Dear Robert Curto:
Thank you for the information.
Warm Regards,
A.R.
Dear Julian Becker:
I do not know, I am not able to make this calculation.
Warm Regards,
A.R.
Dear Mr. Rossi,
I have a question to you and our fellow JONP readers.
How fast do you think a LENR driven rocket could accelerate?
Would it be something quite fast or would it be comparable to an ion-drive or a solar sail.
Just came to my mind. I know it is not related to the Ecat, but I always wanted to know.
Best regards from China,
Julian Becker
Dear Readers, please Google:
ASH TO HYDROGEN GAS
Robert Curto
Ft. Lauderdale, Florida
USA
Mr. Rossi
I hope that you are well. I was curious if there was any news about the delivery of the 1MW plant to your customer/partner. Is it still on track to be delivered by the end of April?
>Otherwise, if you do not publish it, I will be obliged to suit in law the European Physical Journal.
Please keep us updated, Wladimir!
Joe wrote in April 11th, 2013 at 2:48 PM
Wladimir,
In the article that you posted, you mention two forces:
1. Centripetal. Which of the four known forces is this?
2. Spin Interaction. Which of the four known forces is this?
Dear Joe,
1- F= m.V^2/R = m.w.R
2- http://en.wikipedia.org/wiki/Spin%E2%80%93orbit_interaction
regards
wlad
Wladimir,
In the article that you posted, you mention two forces:
1. Centripetal. Which of the four known forces is this?
2. Spin Interaction. Which of the four known forces is this?
All the best,
Joe
Plagiarism in the European Physical Journal
From: Wladimir Guglinski (wladimirguglinski@hotmail.com)
Sent: Thursday, April 11, 2013 2:09:42 PM
To: epja@itkp.uni-bonn.de (epja@itkp.uni-bonn.de)
Cc: jyeston@aaas.org (jyeston@aaas.org); prc@aps.org (prc@aps.org); apr-edoffice@aip.org (apr-edoffice@aip.org); nature@nature.com (nature@nature.com); JOHN ARRINGTON ARGONNE NATIONAL LABORATORY (johna_6@yahoo.com)
Subject: Plagiarism in European Physical Journal
Prof. Ulf Meissner
Editor in Chief
European Physical Journal
Dear Editor,
The European Journal of Physics had published in March 2013 the article The quantum vacuum as the origin of the speed of light , in which is proposed that the space is filled by particles and antiparticles. Such proposal is a plagiarism, because it had been proposed in the article entitled ETHER, published in 2006 in my book Quantum Ring Theory.
But there is a difference between the proposal published now by EPJ in 2013 and my proposal published in 2006, as follows:
a) The authors of the paper published by EPJ had proposed that the space is filled by particles and antiparticles because the new experimental findings published in 2012 require an reevaluation of the concept of space, and so the authors of the paper had proposed it as an ad hoc hypothesis, so that to explain the results of the experiment.
b) Unlike, my proposal in 2006 had been conceived so that to eliminate some inconsistency in the foundations of current Theoretical Physics. So, my theoretical proposal in 2006 actually had represented a prediction to be confirmed by future experimental findings.
In 2012 the journal NATURE had published the article How atomic nuclei cluster, in which there is a plagiarism of a proposal of mine published in 2006 in my book Quantum Ring Theory. Now a new plagiarism is published by European Physical Journal.
The plagiarisms on my ideas have started to occur in reputable journals of Physics (and other plagiarisms will continue to be published) because the new experimental findings are requiring a reevaluation of some current wrong concepts in Theoretical Physics, in order to replace by the wrong concepts by new ones. Since the wrong current concepts had been rejected in my Quantum Ring Theory (and replaced in my theory by new concepts compatible with the new experimental findings published in 2012 and 2013) then it is obvious that any author nowadays (inspired by the results of the new experimental findings) has to proposed the same proposals of mine proposed in 2006, otherwise he cannot explain the new experiments.
In my paper ETHER it is proposed that the space is filled by electric particles e(+), magnetic particles m(+), permeability particles p(+), gravity particles g(+), repulsive particle G(+), and their respective antipariticles e(-), m(-), p(-), g(-), G(-). In the paper it is shown that structure of the space is able to explain the electromagnetic phenomena.
In the paper A MODEL OF PHOTON, published in the page 20 of my book Quantum Ring Theory, it is proposed that the photon is formed by one particle and one antiparticle, they moving with helical trajectory (zitterbewegung). The particle of the photon is composed by the agglutination of the elementary particles of the ether, and the antiparticle of the photon is formed by the elementary antiparticles of the ether. Such model explains all the properties of the light. In the paper it is also shown that from such model we get the Maxwell Equations.
The physicists had used to suppose that the photons are formed by the excitation of the matter (atoms or nuclei) only, that’s why they had used to suppose that the space is an empty vacuum. Now the experiments are showing that photons can be created from the space, which means that the space cannot be an empty vacuum, as they supposed suggested by Einstein’s theory. The new experiments are showing that photons can be created from the structure of the space, and this means that photons are composed from the agglutination of elementary particles and antiparticles of the ether, as proposed in Quantum Ring Theory in 2006.
Dear Prof. Ulf Meissner ,
in order to eliminate the plagiarism commited by EPJ, I suggest you to publish a note in the upcoming issue of the journal , so that to explain to the readers that the hypothesis (regarding a space filled by particles and antiparticles) had been already proposed in my book, in 2006.
Otherwise, if you do not publish it, I will be obliged to suit in law the European Physical Journal.
Regards
Wladimir Guglinski
Dear Gian Luca:
I read the article on “Wired”: honest, sincere, professional.
Warm Regards,
A.R.
Buon giorno A.R.
ho letto l’articolo apparso su “WIRED” di questo mese
http://mag.wired.it/news/2013/04/03/wired-aprile-cyberterrorismo-74545.html
e devo dire che mi ha un pò stupito trovarVi (Lei e il Prof. Focardi) tra quelle pagine. Forsanche perchè siete sempre stati schivi alla grande ribalta ed ai giornali. L’articolo è scritto in maniera appropriata e abbatanza neutra anche nella parte che riguarda la Sua avventura con Petroldragon.
Sarebbe molto bello, quando sarà il momento, conoscere tutti quelli che Le sono stati intorno e che l’hanno aiutata nel lavoro: da F. Fabiani a Majorana.
Spero presto di vederVi sulle prime pagine dei giornali di tutto il mondo.
Cordialità vivissime
Gian Luca
Ps: mi difficile tradurre il testo in inglese e me ne scuso con i lettori.
Dear G. Westreicher:
1- It has been decided by our commercial branch. The issue is complex: you know that in the pioneering phase the prices are higher.
2- no
3- yes
4- confidential
5- obviously there cannot be a definition guaranteed for this issue
Thank you for your insight,
Warm Regards,
A.R.
Dear Mr. Rossi
We learned from you that Your 10KW Ecat home unit will be released to the market after a (undefined?) period of experience with the 1MW industrial plant.
A few thoughts:
At the market introduction of microwave ovens and mobile phones
authorities deliberately accepted, that harmful radiation is emitted,
although no long-term experience was available.
The same also happens very often on pharmaceutical products.
The difference to the Ecat will be probably the missing lobby.
To gain experience quickly, the 1MW plant should be produced to prime costs
and sold in large numbers.
I am aware that the intellectual property of the system in moment is priceless,
but I am convinced, that the production costs are only a fraction of the current selling price.
I`m afraid that with the current strategy the launch of the 1MW plant and the 10KW home units will have a strong delay, and in my view I think this is highly vulnerable.
I have few questions to this topic:
1. ) Why the 1MW plant is so expensive?
2.) Do you (or your background) consciously maintain the 1MW plant so expensive to delay the market launch?
3.) Do you (and your background) have a honest interest in a fast market introduction?
4.) Why there are no campaigns from your “big” Partners to build a lobby for the Ecat?
5.) How long ist the period defined to collect experience from the 1MW plant and to release the 10KW ecat?
I know these are unpleasant questions,
but nonetheless I’m a big fan of You and your technology
and I wish you every success for your project.
Sincerely
G. Westreicher/Austria
Dear Frank Acland:
Leonardo Corporation is still an indipendent company, even after the strategic contract made with our USA Partner.
Warm Regards,
A.R.
Dear Enrico Billi:
Thank you, interesting.
Lavolale, lavolale!
A.R.