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    Quality of Elsevier's Author Support
    Posted on Thursday, March 02, 2017 @ 23:54:18 EST by vlad

    Science 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
    To: wladimirguglinski@_
    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 surveys@elsevier.com
    Yours sincerely,
    Louise Hall
    Market Research

    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

    Dr. 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:

    “My 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 method.
    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:

    <<“Therefore, 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 EST
    (User Info | Send a Message) http://www.zpenergy.com
    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 mathematical test.

    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:

    The 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.

    With the 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.

    1- All  the atomic nuclei have a central 2He4, which produces gravitational rings named n(o)-flux, which captures protons, deuterons, and neutrons.

    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

    5- 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.

    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

    This 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

    The 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.

    The 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 R= Ro.A^(1/3).

    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


    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
    Y= 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.

    When 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 K(Si).
    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:
    K(Si)= 0,614.

    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

    The 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 follows:

    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)

    K(O15)= 0,3142

    The 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

    This 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.

    When the third hexagonal floor is formed, it cancells one of the magnets of the silicon isotopes.
    This means that:
    K(O) = K(Ca)= 1,3715

    Therefore, 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 induction-factor K.

    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)

    In the paper Nr. 7 are calculated several conversions between calcium and oxygen isotopes, and all they give results with good accuracy.

    When 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)

    This 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

    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.


    From 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
    because it is impossible to find any combination of spins from which a null magnetic moment can be obtained, from the foundations of the Standard Nuclear Physics.

    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 nuclear tables.

    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


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