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Quality of Elsevier's Author Support|
Posted on Thursday, March 02, 2017 @ 23:54:18 MST by vlad
WGUGLINSKI writes: Yesterday, 1st March 2017, I received from Elsevier the following email:
From: Elsevier Author Feedback
Sent: Wednesday, March 1, 2017 11:07 AM
Subject: Quality of Elsevier's Author Support
Dear Dr. Guglinski,
I am contacting you because you recently received a final decision on your article submitted to Annals of Physics
We are conducting a short research study to see how satisfied you are with the way your article was managed. Your responses will be used to help improve the publication services we currently offer.
It will only take about 10 minutes to complete the survey online, and your feedback is very important to ensure the accuracy of the research.
If you encounter any problems during the survey, please contact email@example.com
I wrote a series of 8 papers in partnership with Dr. Claudio Nassif, and in the beginning
of 2017 we had submitted the first one to Annals of Physics.
The papers are the following:
Paper Nr. 1: On the reasons why Fermi's theory of beta-decay must be reevaluated
Paper Nr.2: Lorentz factor violation by neutrinos moving with the speed of light
Paper Nr.3: On the origin of mass of the elementary particles
Paper Nr. 4: On how Bohr model of hydrogen atom is connected to nuclear physics
Paper Nr. 5: On how proton radius shrinkage can be connected with Lorentz Factor violation
Paper Nr. 6: Calculation of magnetic moments of light nuclei with number of protons between Z=8 and Z=30
Paper Nr 7: Nuclear spins and calculation of magnetic moments of the isotopes of lithium
Paper Nr 8: Calculation of proton radius to be measured in the Project MUSE
Claudio Nassif is the author of the Symmetric Special Relativity (SSR),
which together with my Quantum Ring Theory compose a Grand Unified
Theory. He has several papers published in the most reputable journals
of Physics worldwide, and his last paper is rated as the second of the Most Read papers of the International Journal of Modern Physics, since 1996:
The first question of the "Quality of Elsevier's Author Support" was concerning my satisfaction with the way my last article in Annals of Physics was managed.
I wrote the following answer:
work is in the brench of Fundametal Physics. From the arguments used by
the Editor for rejecting my paper, I have realized that he has a
personal view on what must be the way for the development of Fundamental
Physics. He neglects the most important experimental findings of the
last 10 years, which deny the some of the fundamental principles of the
Standard Model (SM) and the Standard Nuclear Physics (SNP) , and thereby
he has adopted the strategy of protecting SM and SNP from threatening
experiences which require the reevaluation of some fundamental
principles of those two pillars of Theoretical Physics. Therefore, any
author whose theoretical work does not fit to the personal views of the
Editor, have no chance to be published in Annals of Physics, because the
Editor neglects experimental findings. This procedure of revising and
rejecting articles, based on personal convictions, instead of based on
news experimental findings, is not in agreement with the scientific
Theoretical Physics advancement cannot be subjected to
personal convictions of editors. Theoretical Physics must advance
parallel to the advancement of Experimental Physics.”
Other question was concerning whether my paper was previously submitted to another journal.
My response was “Yes, it was submitted to International Journal of Modern Physics”.
And the next question asked the reasons why my paper was rejected by IJMP.
My response was the following:
“My paper was analyzed by ONLY ONE reviewer. And he used the following anti-scientific argument for rejecting the paper:
the failure of their udd model does not mean we need to abandon
completely the current theoretical paradigm of the nucleon structure,
which is built upon QCD. In other words, they are attacking a theory
that nobody thought was correct”.>>
Then, according to
the referee, the researchers need to continue using the wrong neutron
model ddu, in their search for the discovery of the structure of the
universe, and we have to trust blindly in the discoveries obtained from
such a method of investigation, developed from a model which everybody
know to be wrong.
The criterion used by the referee makes no
sence, because, when we know that a theoretical model is wrong, then
according to the scientifc criterion the theorists have to undertake
efforts in order to discover a better model.
Finally, Elsevier asked me to give any suggestion, for the aim of improving the quality of their publications.
And I sent the following reply:
“The Editors must respect the scientific method. The scientific experiments are more important than their personal opinions”.
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|New nuclear model of Hexagonal Floor under mathematical test (Score: 1)|
by vlad on Sunday, December 10, 2017 @ 17:31:06 MST
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|WGUGLINSKI writes: The new nuclear Hexagonal Floors model, proposed in my book Quantum Ring
Theory, published in 2006 by the Bauu Institute Press, is under
The test is under review through the three following papers:
Paper Nr. 6: Calculation of magnetic moments for light nuclei between Z=8 and Z=30 - PART ONE
Paper Nr. 7: Testing the equations of the new nuclear Hexagonal Floors model
Paper Nr. 9:
first version of the Paper Nr. 6 was rejected by the International
Journal of Modern Physics E, in 2016, because at that time I did not
discover yet an anomoly of the silicon isotopes.
discovery of the silicon isotopes, the new version of the Paper Nr. 6
was submitted to the Pramana Journal of Physics, in August 2017,
together with the Papers Nrs. 7 and 9.
As the first version of the
paper was rejected by the IJMP-E in 2016, the new version was submitted
again to the journal, together with the Papers Nrs. 7 and 9.
Ahead it is given an idea on how the mathematical test of the new nuclear model is going on.
All the atomic nuclei have a central 2He4, which produces
gravitational rings named n(o)-flux, which captures protons, deuterons,
2- The atomic nuclei are divided in two sides: Ana
and Douglas, and they constitute two magnetic poles of the nucleus:
Ana is south, and Douglas is north.
3- All the atomic nuclei rotate at the ground state.
4- The isotope oxygen-16 is formed by a central 2He4, and its n(o)-flux captures six deuterons, which form an hexagonal floor
As O16 rotates in the ground state, one had to expect that the six
charges of the six protons (of the deuterons) would induce a magnetic
moment, due to their rotation, and therefore O16 could not have null
magnetic moment. However, as 3 deuterons are in the south pole, and 3
deuterons are in the north pole, their total contribution for the
induction of magnetic moment is zero.
6- INDUCTION FACTOR "K":
The isotope oxygen-15 is formed by five deuterons and one proton moving about the central 2He4.
From the structure of oxygen-15, we get:
The magnetic moment due to the proton and the unpaired deuteron, without considering the rotation of the oxygen-15, is:
-2,793 + 0,857
The rotation of the oxygen-15 induces the following additional magnetic moment:
The experimental magnetic moment of oxygen-15 is 0,719.
-2,793 +0,857 +K.(+2,793-0,857) = 0,719
K = 1,3715
value K= 1,3715 is the induction factor in the oxygen-15 isotope. It
is caused by the rotation of the six charges of the six protons moving
about the central 2He4.
This means that each proton is responsible for the following induction factor:
K(p) = 1,3715^(1/6), for each proton
7- POWER ROTATION PwR:
rotation of atomic nuclei in the ground state is caused by the spin of
the protons, neutrons, and deuterons which move about the central 2He4.
The intensity of the rotation is proportional to the intrinsic magnetic moment of the nucleon.
For instance, the PwR induced by a neutron is proportional to 1,913
The PwR induced by a deuteron is proportional to 0,857
The PwR induced by a proton is proportional to 2,793
The direction of the rotation depends on the direction of the spin.
The PwR of a nucleus is calculated as follows:
PwR = [somatory of contribution of all neutrons, protons, and deuterons] divided by the inertia moment of the isotope.
inertia moment of the isotope is given by A.R^2 , where A is the atomic
mass of the isotope, and R is its radius given by the empirical formula
The radius of oxygen-15 is:
R= 1,25x15^(1/3) = 3,08277
The calculation gives for the PwR(O15) of oxygen-15:
PwR(O15) = 0,03108
The PwR gives the angular velocity of the isotope
8- THE FUNDAMENTAL EQUATION
For one isotope X, its magnetic moment m(X) is calculated by the following equation:
m(X) = 1,3715^(Y/6).[ PwR(X).R(X) ] / PwR(O15).R(O15) ]
1,3715 = induction factor of oxygen-15
number of proton charges moving about the central 2He4. For instance,
for lithium isotopes Y=1, for Be isotopes Y= 2, for B isotopes Y= 3, for
C isotopes Y = 4, for N isotopes Y= 5.
PwR(X) = power rotation of the isotope X
R(X)= power rotation of the isotope X
PwR(O15) = power rotation of the oxygen-15
R(O15)= power rotation of the oxygen-15
With the FUNDAMENTAL EQUATION are calculated the magnetic moments for several isotopes, between Z=1 and Z=7.
9- THE ANOMALY OF SILICON ISOTOPES
the second hexagonal floor is completed in the silicon isotopes, due to
the magnetic repulsions between the south and north poles, the second
floor gyrates by 180º.
Therefore, while oxygen isotopes have
only one magnet responsible for the production of their induction factor
K(O), in the silicon isotopes there are TWO magnets responsible for
So, we have to expect that K(Si) = 2xK(O).
The induction factor for excited Si28 with spin 2+ is calculated, and gives the result:
The radius of silicon-28 is:
R= 1,25x28^(1/3) = 3,7957
The calculation gives for the PwR(exSi28) of excited silicon-28:
PwR(exSi28) = 0,005782
conversion of the induction factor K(O15)= 1,3715 for the conditions of
angular velocity and radius of the silicon-28 can be calculated as
If oxygen-15 had a radius R(O15)= R(Si28)= 3,7957 and
also a power rotation PwR(O15)= PwR(exSi28)= 0,005782, then the value of
the induction-factor K(O15) would be:
K(O15) = 1,3715x0,005782x3,7957 / (0,03108x3,0827)
result obtained shows that, if oxygen-15 was working in the same
conditions of exSi28, then the ratio K(exSi28) / K(O15 would be:
0,614 / 0,3142 = 2,041
result confirms what we have expected: the induction-factor of silicon
isotopes is twice of that for oxygen isotpoes, because silicon isotopes
have TWO MAGNETS, while oxygen isotopes have ONLY ONE magnet.
10- INDUCTION FACTOR FOR CALCIUM ISOTOPES
When the third hexagonal floor is formed, it cancells one of the magnets of the silicon isotopes.
This means that:
WE HAVE TO EXPECT THAT OXIGEN AND CALCIUM ISOTOPES HAVE THE SAME VALUE FOR THE INDUCTION FACTOR:
K(O) = K(Ca)= 1,3715
it is possible to convert several calcium isotopes to the conditions of
the oxygen-15 isotope, so that to verify if they have the same
The first conversion was made between oxygen-15 and calcium-39, because both them have an unpaired proton in their structure.
The calculus, made in the Paper Nr. 6, obtained the foloowing value for the conversion:
Conv [ K(Ca39) => K(O15) ] = 1,371144
having a difference 0,000353 regarding the value 1,3715 = K(O15)
the paper Nr. 7 are calculated several conversions between calcium and
oxygen isotopes, and all they give results with good accuracy.
11- INDUCTION FACTOR FOR IRON ISOTOPES
the fourth hexagonal floor is completed in the iron isotopes, there is
formation of TWO magnets again, and therefore we have to expect that
silicon and iron isotopes have the same induction factor:
K(Si) = K(Fe) = 2 x 1,3715 = K(O) = K(Ca)
is tested in the Paper Nr. 9, and the results have confirmed that iron
and silicon isotopes have the same induction-factor, which is twice of
that for oxygen isotopes
12- The FUNDAMENTAL EQUATION:
m(X) = 1,3715^(Y/6).[ PwR(X).R(X) ] / PwR(O15).R(O15) ]
was used for the calculation of the magnetic moments of several isotopes: Li, Be, B, C, N, F, Mg, Ar, etc.
All the calculation have given good accuracy with experimental data quoted in nuclear tables.
13- THE IMPOSSIBLE NULL MAGNETIC MOMENT FOR EXCITED ISOTOPES WITH PAIR AND EQUAL NUMBER OF PROTONS AND NEUTROS.
the foundations of the Standard Nuclear Physics it is IMPOSSIBLE to
explain why is NULL the magnetic moment of the excited isotopes with
spin 2+, with equal number of protons an neutrons, Z=N:
C12, O16, Mg24, Ar36, Ca40
it is impossible to find any combination of spins from which a null
magnetic moment can be obtained, from the foundations of the Standard
Excited Ca42 with spin 2+ also has null magnetic
moment, but his electric quadrupole moment is quoted in nuclear tables,
and therefore the experimentalist have tried to measure its magnetic
moment, but it is not quoted in nuclear tables, and the reason is
because its magnetic moment is zero.
The same happens with other
isotopes with Z and N pairs. Their quadrupole moments is quoted, but
not their magnetic moments, and therefore the epxerimentalists did not
report the value zero of their experiments for the editors of the
The reason why those isotopes have null magnetic moment is shown in the Paper Nr. 7.
The editors of the International Journal of Modern Physics E, and Pramana Journal of Physics, are in silence since August 2017.
Let us wait what they say about the matemathical test for the Hexagonal Floors model
|Calculation of proton radius to be measured in the Project MUSE (Score: 1)|
by vlad on Thursday, April 05, 2018 @ 20:26:28 MST
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|Submitted by WGUGLINSKI:|
Published in Physics Essays (30 March 2018):
to the Standard Model (SM), the proton must be unshrinkable, and the
two different values 0.8758 and 0.8770 fm (respectively, measured using
the atomic hydrogen method and proton-electron scattering) must be
related to a unique average value of 0.8860.01 fm, with discrepancies
due to errors that are inherent in each of the two methods. However, as
shown here, there are strong reasons for assuming that the two values
0.8758 and 0.8770 fm actually come from two different physical
mechanisms that are responsible for proton radius shrinkage. In the
hydrogen atom method, the cause is the electric charge of the electron;
in the proton-electron scattering method, the cause is the mass of the
electron. In 2010, measurements of energy levels of muonic hydrogen
suggested a proton size of 0.84260.001 fm, which was confirmed by
experiments in 2013. The “MUon proton Scattering Experiment” (MUSE) is
an effort to expand the comparisons by determining the proton radius
through muon scattering with simultaneous electron scattering
measurements. Here, it is calculated that the proton radius that will be
measured via muon scattering in the MUSE project (to be conducted
between 2018 and 2019) will be between 0.616 and 0.722 fm.
Key words: Proton
Radius Shrinkage inside Atomic Nuclei; Proton Radius to be Measured in
Project MUSE; Discrepancy between the Neutron Lifetime Measured in Beam
and Bottle Experiments.
|New procedure to be tested in JLab for neutron radial charge distribution (Score: 1)|
by vlad on Wednesday, April 25, 2018 @ 19:07:51 MST
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Submitted by WGUGLINSKI:
Dear Dr. Adrian Cho
Staff Writer, Science
I did not find the email of Dr. Stuart Henderson, Director of DOE's Jefferson Lab, and so I would like to ask you the favour to send him this email.
In June/2018 the International Journal of Fundamental Physical Sciences will publish a paper of mine, in which it's proposed an interesting experiment with the aim to measure the neutron radial charge distribution , but with a new procedure different of that already made in Jefferson Lab, in 2007.
The ABSTRACT of the paper is the following:
In another author’s paper is proposed that proton and neutron radii have contraction inside the atomic nuclei (Guglinski, 2018). In that paper it’s also proposed that the discrepancy of 8s, between the neutron lifetime measured in beam and bottle experiments, is caused by the contraction of the neutron radius inside atomic nuclei. According to the present theory, the neutron radius in beam experiments dilates from 0,26fm up to 0,87fm, and after 8s of the dilation begins the process of decay. Here is proposed a new model of neutron with quark structure d(u-e-u), where an electron is sandwiched between two up quarks. From the shell thickness 2b=1,10 fm of charge distribution in atomic nuclei, it is obtained that the orbit radius of the electron inside the neutron is Re= 0,26fm. Starting up from this value, it’s calculated the value of the electric quadrupole of the deuteron, and the theoretical value is the same measured by experiments: 2,7.10^-31 m² . The value Re=0,26fm, of the electron orbit, is used for the calculation of the neutron radial charge distribution, but the result does not reproduce exactly the distribution measured by Jefferson Lab. In order to replicate theoretically with good accuracy the charge distribution measured by JLab, there is need to consider a value Re=0,31fm. As the neutron has shrinkage inside the structure of the deuteron, where the neutron is no free (but the experiments in JLab use free neutrons), then it’s reasonable to consider that the orbit radius of the electron in the neutron of a deuteron is 0,05fm shorter than that in a free neutron. If this difference of 0,05fm really exists, then neutrons used in beam and bottle experiments have a little difference in their radial charge distribution (during the 8 seconds of the dilation of the neutron radius). The hypothesis of the existence of such difference can be verified by measuring, in JLab, the radial distribution of neutrons used in beam experiments, but under a new condition: the distribution must be measured in the first initial 8s of the neutron decay, along which occurs the dilation or the radius Re=0,26fm. If detected a little difference, regarding the data already collected by JLab for the neutrons used in the beam experiments, then the model of neutron d(u-e-u) will be confirmed, as also the contraction of the proton and neutron inside atomic nuclei. A procedure for the test in JLab is proposed in the paper.
Fermi’s beta-decay, New neutron quark model, Neutron radial charge distribution, New procedure to be tested in Jefferson Lab.
REFERENCE quoted in the Abstract:
Guglinski, W (2018). Calculation of proton radius to be measured in the Project MUSE. Physics Essays, p 137.
As the experiments have detected a discrepancy of 8s in the beam and bottle experiments, I think there is a good reason to verify wheter a difference will appear also in experiments measuring the neutron radial charge distribution.
So, please send this email to Dr. Henderson.
Thanks to your attention
|Theoretical calculation of the neutron mag. moment from 2 independent procedures (Score: 1)|
by vlad on Tuesday, June 05, 2018 @ 19:37:04 MST
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|WGUGLINSKI writes: The International Journal of Fundamental Physical Sciences published in June-2017 the paper “Re-evaluation of Fermi’s theory of beta-decay”:|
In the paper it is proposed a new quark model for the neutron, whose structure is d(u-e-u), where an electron is sandwiched between two up quarks.
The magnetic moment of the neutron model d(u-e-u) is calculated from two different and independent methods of calculations, and they give two very close values: 1,9071 and 1,9073, both them close to the experimental value 1,913.
There is NOT mathematical connection between the two methods, and therefore the two successful calculations cannot be considered as a “mathematical coincidence”.
The two independent methods are described ahead.
1- In this method is NOT used the velocity of the particles. Therefore it is NOT used the well-known equation m= -(e/2m).L.
2- In the proton the two up quarks have orbit radius R(u)=0,45fm, whereas the down quark has R(d)=0,80fm, measured in 2007 at the Jefferson Laboratory.
3- In the neutron model d(u-e-u), the orbit radius of the electron is Re= 0.31fm, calculated from the electric quadrupole moment of the deuteron, as shown in the equation 2 of the paper.
4- In the theoretical neutron model d(u-e-8) the orbit radii of the up quarks u’ and u’’ are R(u’) = 0,28fm and R(u’’) = 0,42fm. Whereas the orbit radius of the down quark is Rd=0,80fm.
5- Considering the orbit radii are Re=0,31fm , R(u’)=0,28fm , R(u’’)=0,42fm , Rd=0,80fm , the neutron model d(u-e-u) reproduces very well the neutron radial charge distribution measured by the Jefferson Lab in 2007, as seen in the graphic shown in the Figure 28 of the paper. Although the radial charge distribution measured by JLab was very well reproduced by the theoretical model d(u-e-u), nevertheless the confirmation of this model requires to verify if is able to reproduce the magnetic moment of the neutron, m=1,913. The first method of calculation was the following:
6- From the radii Ru=0,45 and Rd=0,80, and the charges (+2/3).e of each up quark and (-1/3).e of the down quark, it is calculated the percentage Y=69,23% due to the contribution of the two up quarks, and Z=30,77% due to the contribution of the down quark, as seen in the equations 38 and 39 of the paper.
7- Starting from the experimental magnetic moment m(p)= +2,793 of the proton, it is calculated how the magnetic moment due to the up quarks varies which the variation of the orbit radii R(u’) and R(u’’) of each up quark, as shown in the equation 40.
8- The calculation of the magnetic moment produced by the neutron model d(u-e-u) is calculated from the equation 41 up to 45. The value calculated is 1,9071.
1- In this method it’s used the velocity of the particles, and the well-known equation m= -(e/2m).L. And therefore there is not mathematical connection between the two methods of calculation.
2- Again, the orbit radii in the neutron are Re=0,31fm , R(u’)=0,28fm , R(u’’)=0,42fm , Rd=0,80fm
3- The velocity of the up and down quarks in the d(u-e-u) model is calculated with the equation m= -(e/2m).L, as explained in the items 4 and 5 ahead.
4- The velocity of the quarks up and down is obtained from the experimental proton magnetic moment m(p)= +2,793 (see equation 46 of the paper). These velocities of the quarks, calculated for the proton, are applied in the neutron model d(u-e-u).
5- The velocity of the electron in the model d(u-e-u) is obtained from the Kurie’s graphic for the beta-decay of neutron (see equation 13).
6- The magnetic moment of the neutron model d(u-e-u) is calculated in the equation 47, giving m(n)= -1,9073.
1) As seen, the two methods are independent from the MATHEMATICAL viewpoint, because the first method uses the velocity of the particles, and the second method does not use it.
2) Therefore, the fact that the two values [ m(n)= -1,9071 and m(n)= -1,9073 ] are very close must be credited to the following reasonable argument:
3) There is a physical connection between the THEORETICAL neutron model d(u-e-u) and the PHYSICAL neutron model existing in the nature.