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    Quality of Elsevier's Author Support
    Posted on Thursday, March 02, 2017 @ 23:54:18 GMT 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 GMT
    (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

    Calculation of proton radius to be measured in the Project MUSE (Score: 1)
    by vlad on Thursday, April 05, 2018 @ 20:26:28 GMT
    (User Info | Send a Message) http://www.zpenergy.com
    Submitted by WGUGLINSKI:

    Published in Physics Essays (30 March 2018):

    According 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 GMT
    (User Info | Send a Message) http://www.zpenergy.com
    Submitted by WGUGLINSKI:
    Tue 4/24/2018, 11:49 PM

    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.

    Key words:

    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
    W Guglinski

    Theoretical calculation of the neutron mag. moment from 2 independent procedures (Score: 1)
    by vlad on Tuesday, June 05, 2018 @ 19:37:04 GMT
    (User Info | Send a Message) http://www.zpenergy.com
    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.


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