BCC lattice model of nuclear structure

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by
Gamal A. Nasser
Faculty of science, Mansoura University, Egypt
E-mail: chem.gamal@hotmail.com
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Abstract
This model is development of solid nuclear models. Like FCC model, this model can account for nuclear properties that have been explained by different models. This model gives more accurate explanation for some nuclear properties which are Asymmetric fission, Nuclear binding energy and the most bound nuclei, Natural radioactivity and Number of neutrons in nuclei depending on the structures of these nuclei. The structures of nuclei in this model have special advantage, as there is separation between lattice positions of similar nucleons giving new concept for nuclear force.
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565 comments to BCC lattice model of nuclear structure

  • bertoldo

    https://www.facebook.com/video.php?v=10152540866371083&set=vb.215052536082&type=2&theater Di questo video cosa ne pensa Dott. Rossi . minuto 0:25 – 0:30 . L’utilizzo delle frequenze ha questi risultati anche nella sua macchina ?

  • Joe

    JR,

    I thank you for your response.

    Two more points if I may:

    1. An example of a particle property is mass. This property pre-exists within the Schrodinger equation in the variable m. Therefore the need to derive this property by way of the Schrodinger equation is voided. If particle nature is truly ultimately wave nature, the Schrodinger equation should be able to derive mass instead of pre-establishing it within its structure. But it can not derive it. And I contend that the reason is that particle nature is foreign to the Schrodinger equation.

    2. Why would the wave function collapse only when we humans interact with it (by way of measurement, etc) and not when anything else in the Universe interacts with it, which actually happens at every moment? What if many humans looked for the object simultaneously, should not the collapse of the wave function happen in many places accordingly and all at once, or maybe not happen at all?

    All the best,
    Joe

  • James Heath

    Dear Mr Rossi,
    Without giving any proprietary information can you answer my following question? How important is the atmosphere of the reaction, is a specific amount of multiple elements needed or just a certain amount a specific one? Do you believe the reaction could be versatile enough for it to occur suspended in a liquid with the right catalysts present?

  • Wladimir Guglinski

    The mystery on the electric quadrupole moment for the deuteron

    Dear readers of the JoNP.

    In order to understand why from the Standard Nuclear Physics is impossible to get the electric quadrupole moment Q(b) of the deuteron, first of all we have to understand what is elec. quadr. moment Q(b).

    Electric quadrupole momnent is concerning the distribution of electric charges.
    We have:
    1) An elipsoidal distribution of charges has Q(b) different of zero
    2) A spherical distribution of charges Q(b) = 0.

    The deuteron is formed by proton+neutron.
    The proton has a spherical distribution of charge, and therefore it has Q(b)=0.
    The neutron has charge zero, and therefore the neutron has Q(b)=0.

    So, a deuteron formed by p+n would have to have elec. quad. mom. Q(b)=0.

    Therefore one of the biggest and harder dramas of the Standard Nuclear Physics was born in 1939 when the physicists discovered that deuteron has non null electric quadrupole moment, as expected from the Standard Nuclear Physics, but actually it has Q(b) = +0,028

    The paper was published in the Jornal Nature:
    The Electric Quadrupole Moment of the Deuteron and the Field Theory of Nuclear Forces
    http://www.nature.com/nature/journal/v144/n3645/abs/144476a0.html
    —————————————————————-
    “THE discovery by Rabi and his collaborators that the deuteron in its ground state possesses an electric quadrupole moment is of considerable theoretical importance, since it clearly shows that the forces acting between a proton and a neutron must to a quite appreciable extent depend on the spatial orientations of the spins of the heavy particles.”
    ——————————————————————-

    Along 66 years the theorists tried to solve the mystery.
    But of course they did never succeed, because the model of neutron considered in the Standard Model is wrong.

    In 2005 (therefore 66 years after the publication of the paper by Nature in 1939), the puzzle of the nucleon-nucleon force was not solved yet, as we realize from the paper published by the Tukish Journal of Physics:

    Electron-Deuteron Tensor Polarization and D-State Probability
    http://journals.tubitak.gov.tr/physics/issues/fiz-05-29-3/fiz-29-3-1-0408-3.pdf
    Page 129:
    —————————————————————-
    ”One of the main hopes of electron-deuteron scattering experiments have been to measure certain features of the deuteron wave function and to use these properties to determine unknown properties of the nucleon-nucleon force”.
    —————————————————————-

    Thirty three models were proposed along 65 years, as we see in the page 129 of the paper published in the Turkish Journal of Physics:
    —————————————————————-
    ”So, in our investigation of deuteron tensor polarization we employ thirty-three local potential models of the nucleon-nucleon force.”
    […]
    “These thirty-three potential models have different deuteron properties, such as deuteron quadrupole moment Q , D-state probability P , asymptotic D-state amplitude A and asymptotic ratio E. The values of these properties are not equal, but have wide range of values in all potential models.”
    […]
    ”To discuss the properties of various wave functions of these potential models, the deuteron radial wave functions u and w of fourteen selected local potential models among the above mentioned 33 potential models (i, c, f, GK3, TSC, r6, RSC, RHC, r7, HJ, PARIS, MHKZ, 2 and 4) are chosen. “
    —————————————————————-

    Of course the nuclear theorists may continue to try to solve the puzzle of the elecrric quadrupole moment of the deuteron along 660 years, or 6600 years, or 66000 years, but they will never succeed to solve the puzzle if they continue keeping the model of neutron considered in the Standard Model. They can propose 330 models, 3300 models, or 33000 models, and they will never succeed to solve the puzzle.

    There is only one way to solve the puzzle: it is by considering the new model of neutron formed by proton + electron, n=p+e, proposed in Quantum Ring Theory.

    The electric quadrupole moment Q(b) of the deuteron is calculated in my paper Anomalous Mass of the Neutron, published in the JoNP.
    See page 43, equation 15:
    http://www.journal-of-nuclear-physics.com/files/Anomalous%20mass%20of%20the%20neutron.pdf

    The theoretical value calculated in the paper is Q(b)= 3×10^-31m² , while the experiments get Q(b)= +2,86×10^-31m².

    My paper Anomalous Mass of the Neutron was submitted to the Chinese Journal of Physics in 2002, and the referee rejected the paper because in my paper it is considered that the radius of the proton is R= 0,275fm, while in the Standard Model the proton’s radius is R= 0,8fm (obtained by experiments via scattering proton-electron).

    However, in 2011 new experiments, made with a different method, had measured the proton’s radius to be a little shorter than R=0,8fm.

    And the question now is: is that result either due to errors in the measurement or due to the fact that the proton’s radius can be shorter than 0,8fm.

    My opinion is that proton’s radius has a shrinkage when it is bound with other particles heavier than the electron.. A free proton has radius R=0,8fm (as measured in the scattering proton-electron experiments), but when it is bound with heavier particles its radius is very shorter than 0,8fm

    In order to eliminate the controversy on the proton’s radius, in 2015-2016 other experiment will be made so that to measure the proton’s radius via scattering of the proton with mesons.
    As the meson has mass 200 times heavier than the electron, I expect that the new experiments to be made between 2015 and 2016 will measure the proton’s radius very shorter than 0,8fm (probably between 0,3fm and 0,6fm).

    Now we have to wait the results of the experiments

    regards
    wlad

  • Andrea Rossi

    Pekka Janhunen:
    Thank you for your insight.
    Warm Regards,
    A.R.

  • Dear Andrea Rossi,

    Some ideas for possible LENR theories. Maybe the effective electron mass becomes zero or very small for some reason. This is not principally impossible, since a similar thing happens in graphene. Then the electron plasma frequency which is proportional to sqrt(n/m) [n is electron density, m is effective electron mass] becomes high. In other words, electron plasma oscillations (plasmons) acquire high energy. Maybe such high-energy plasmons can then couple to nuclear degrees of freedom and explain why no radiation is emitted. Maybe this also helps explain why reactions happen in the first place, by making new reaction exit channels available.

    Some weeks ago I suggested to you an experiment where a gamma ray source is put behind the reactor and then one checks if the ability of the reactor to suppress and/or scatter the gamma rays increases when the reactor is turned on. If it would happen, it would be consistent with the above picture i.e. presence of energetic plasmons.

    Some people have earlier pondered about the possibility of increasing the effective electron mass and thereby to more easily overcome the Coulomb barrier, i.e., a similar thing which happens in muon catalysed fusion. However, in that case I don’t know how radiation would be suppressed. My suggestion above is the reverse: instead of making the electron mass higher, consider what happens if one makes it lower. Then the agents to enable the reaction wouldn’t be electrons (as particles), but plasmons (as collective oscillation modes of the electron fluid).

    To explain LENR (chemical environment assisted nuclear reactions), somehow one has to build a bridge from low-energy chemistry to high-energy nuclear phenomena. All particles in chemical systems have low energy, so the task appears difficult. But maybe some collective wave modes (such as plasmons) can have high energy. It is a possible loophole that looks interesting to me.

    regards, /pekka

  • Andrea Rossi

    Tom Conover:
    Sharing my thoughts in this blog is an enrichment for me ( and a precious source of corrections too). I hope the same for our dear Readers.
    Warm Regards,
    A.R.

  • Tom Conover

    Dear Andrea Rossi,

    For readers seaching for the replication of the death ray used against the fleet of the enemies assailing Syracuse using mirrors that reflected the sunrays that you referenced perhaps they can see the flames created at:

    http://web.mit.edu/2.009/www/experiments/deathray/10_ArchimedesResult.html

    Too bad M.I.T. couldn’t replecate the Pons and Fleishman experiment in 1989 – eh? Thank you for sharing your thoughts about your work with us on this web site, Andrea!

    Tom Conover

  • Wladimir Guglinski

    REPLY Nr. SEVEN

    Ab initio calculation of energies of light nuclei with the Hybrid Multideterminant scheme
    http://link.springer.com/article/10.1140%2Fepja%2Fi2006-10214-6
    Abstract.
    We use the AV8′ nucleon-nucleon potential renormalized with the Lee-Suzuki prescription with the Hybrid Multideterminant scheme to evaluate energies of some light nuclei. The Lee-Suzuki prescription is used to evaluate the the two-body matrix elements up to 6 major oscillator shells in the lab frame. The Hybrid Multideterminant scheme is used to deal with the nuclear-structure problem. The results obtained for 6Li, 12C and 16O are compared with the results obtained with other methods. The results suggest a reasonable convergence of the renormalization prescription for 6 major shells.
    =================================

    COMMENT

    1) Ab initio ???
    Actually ab initio would be to solve the primordial puzzle on the nucleon-nucleon force, not solved yet.

    2) Nothing concerning why 8Be is unstable.

    3) However, we see one more method, the Hybrid Multideterminant sheme, and so we go back to what we said in the REPLY Nr. ONE: Why so many methods ???

    The puzzle of the nucleon-nucleon force in the deuteron (the most simples nucleon formed by p+n) was not solved yet, in spite of along 66 years 33 theories were proposed.

    But in the Abstract the authors say:
    We use the AV8′ nucleon-nucleon potential.
    First of all, we realize that the AV8’ nucleon-nucleon potential is not quoted in the paper published by the Turkish Journal of Physics. So, while those 33 nucleon-nucleon potential mentioned in the TJP were conceived so that to satisfy the nuclear properties of the nucleon-nucleon potential existing in the deuteron, we don’t know if the AV8 was conceived in order to consider also the puzzles of the deuteron.

    4) So, a fundamental question arises: by considering the principles of the Standard Model, as the primordial puzzle of the nucleon-nucleon force in the deuteron (formed by proton+neutron) was not solved yet, then does it make sense to hope to solve the puzzles of the light nuclei ??? (since they are composed by proton+neutron).

    .

    Finally,
    I would like to say that I am very thankful to you, Mr. JR, because you promoted to me the opportunity to show how the current nuclear theories are full of unsolved puzzles.

    Regards
    wlad

  • Wladimir Guglinski

    REPLY Nr. SIX

    Energy levels of light nuclei A = 5−10
    http://www.sciencedirect.com/science/article/pii/0375947488901248
    Abstract
    A review of the evidence on the properties of the nuclei with A = 5, 6, 7, 8, 9 and 10, with emphasis on material leading to information about the structure of the A = 5−10 systems.
    ===================================

    COMMENT:

    Nothing concerning why 8Be is unstable.

    regards
    wlad

  • Wladimir Guglinski

    REPLY Nr. FIVE

    Alpha decay constant of 8Be nucleus
    http://www.worldscientific.com/doi/abs/10.1142/S0217732314500278
    The 8Be nucleus decays into two 4He nuclei. This decay constant is theoretically estimated using Fermi golden rule and the ground state wave functions of the 4He and 8Be nuclei. The estimated result agrees pretty well with the reported experimental value.

    .
    Fermi’s golden rule
    In quantum physics, Fermi’s golden rule is a way to calculate the transition rate (probability of transition per unit time) from one energy eigenstate of a quantum system into another energy eigenstate, due to a perturbation.
    http://en.wikipedia.org/wiki/Fermi%27s_golden_rule
    ================================================

    .

    COMMENT
    So, it is not a theory, based on a nuclear model, and it does not explain why 8Be is unstable.
    It is actually a phenomenological work, based on the experimental fact that 8Be nucleus decays into two 4He nuclei, and uses the Fermi’s golden rule (and such rule has nothing to do with the stability of the nuclei).

    The nucleus 8Be has a binding energy of 7,06MeV/nucleon, however the nucleus is unstable. The paper calculates the binding energy, but does not explains why 8Be is unstable.

    Besides, as already mentioned in the REPLY Nr. THREE, the 8Be formed by two-alpha clusters is not viable, because the 9Be formed by 8Be+n cannot reproduce the nuclear properties of the 9Be.

    regards
    wlad

  • Wladimir Guglinski

    REPLY Nr. FOUR

    Quantum Monte Carlo calculations of A=8 nuclei
    http://journals.aps.org/prc/abstract/10.1103/PhysRevC.62.014001
    Abstract
    We report quantum Monte Carlo calculations of ground and low-lying excited states for A=8 nuclei using a realistic Hamiltonian containing the Argonne v18 two-nucleon and Urbana IX three-nucleon potentials. The calculations begin with correlated eight-body wave functions that have a filled α-like core and four p-shell nucleons LS coupled to the appropriate (Jπ;T) quantum numbers for the state of interest. After optimization, these variational wave functions are used as input to a Green’s function Monte Carlo calculation made with a new constrained path algorithm. We find that the Hamiltonian produces a 8Be ground state that is within 2 MeV of the experimental resonance, but the other eight-body energies are progressively worse as the neutron-proton asymmetry increases. The 8Li ground state is stable against breakup into subclusters, but the 8He ground state is not. The excited state spectra are in fair agreement with experiment, with both the single-particle behavior of 8He and 8Li and the collective rotational behavior of 8Be being reproduced. We also examine energy differences in the T=1,2 isomultiplets and isospin-mixing matrix elements in the excited states of 8Be. Finally, we present densities, momentum distributions, and studies of the intrinsic shapes of these nuclei, with 8Be exhibiting a definite 2α cluster structure.
    ===========================================

    COMMENTS:

    1) The filled α-like core and four p-shell nucleons is similar to the nuclear model proposed in Quantum Ring Theory, since in QRT the nuclei have a central 2He4 (core).

    2) The definite 2-alpha cluster structure for the 8Be is no viable, as explained in the REPLY No. THREE, because a model of 9Be formed by 2-alpha cluster+neutron cannot reproduce the magnetic moment, nuclear spin, and electric quadrupole moment exhibited by the 9Be , as measured by experiments.

    regards
    wlad

  • Wladimir Guglinski

    REPLY Nr. THREE

    The Very Rich Structure of the Rather Light Nuclei
    http://www.cenbg.in2p3.fr/heberge/EcoleJoliotCurie/coursannee/cours/W-Catford.pdf
    =======================================
    6. MASS A=8
    The 8Be system is, in some ways, the most enigmatic of all the beryllium isotopes.
    It can be considered as two alpha-particles in orbit with zero angular momentum.
    ========================================

    .

    COMMENT 1:

    Unfortunatelly, the 8Be has not nuclear properties so that to verify if such structure formed by apha-particles is satisfactory.

    However, we can verify the alpha-particles model by considering the 9Be. Look at what is said in the item 7, MASS = 9:
    ======================================
    The molecular description of 9Be, in terms of two alpha-particles bound together by a neutron in a molecular orbital, was developed by Seya and collaborators in 1981 [21].
    ======================================

    .

    COMMENT 2:
    This model is impossible. Each alpha-particle has nuclear spin zero, and therefore the 9Be would have to have the spin of the neutron, i=1/2 . However, from experiments we know that 9Be has spin 3/2.

    Other problem is the magnetic moment. Each alpha-particle has magnetic moment zero. So, the magnetic moment of the 9Be is due to the neutron, and its magnetic moment is -1,913. But due to the rotation of the nucleus, the magnetic moment has to increase of about 10%, and therefore 9Be could not have magnetic moment less than -2,10. But the experiments show that 9Be has magnetic moment -1,177.

    Other problem is the quadrupole moment. The two alpha-particles produce null quadrupole moment. As the neutron has no electric charge, the electric quadrupole moment of the 9Be must be zero. However the experiments detected that 9Be has electric quadrupole moment +0,053.

    .

    COMMENT 3:

    In the paper it is written:
    ======================================
    Although the alpha-particle subsytems bring with them a high binding energy, the 8Be system is not bound overall.
    ======================================

    So,
    in spite of they calculate the binding energy of the 8Be, however it is not explained why the 8Be is unstable.

    Actually only a conjecture is proposed:

    ===============================
    The existence of a loosely bound system of identical boson particles (alpha-particles) could lead one to suppose [15] that a link might exist with Bose-Einstein condensates.
    =================================

    But Bose-Einstein condensates refers to a gas of bosons. So, by considering 8Be as cluster of two bosons 2He4, an the oxygen 16.O a clusters of 4 bosons 2He4, a question arises: Why 8Be is unstable, and 16.O is stable? They both are formed by a pair number of bosons.
    So, by considering the conjecture of the Bose-Einstein condensate, actually 16.O would have a structure formed by two 8Be nuclei, and therefore 16.O would be unstable like the 8Be is.

    However, as the theorists know that 16.O is stable, of course they invent a mathematical artifice, in order to explain why a Bose-Einstein condensate formed by 4-boson nuclei 2He4 is stable, by following the example given by Heisenberg, when he invented the mathematical artifice named Isospin, so that to explain why two neutrons are not bound via strong nuclear force.
    When a theory fails to give what the logic expects from it, the solution is via mathematics, by inventing mathematical phantasmagoric assumptions.

    .

    COMMENT 4:

    In the paper it is written:
    ===================================
    Another interesting feature of nuclear structure is illustrated by the unbound ground state of 8Be, and is sometimes referred to as the phenomenon of
    “ghosts” [18].
    ===================================

    Well,
    It is not a surprise.
    Because the Standar Nuclear Physics was developed from the Heisenberg’s phantasmagoric scientific method.

    And what we can expect from a phantamagoric method, if not ghosts?

    .

    COMMENT 5:

    The paper is ended with the following words:
    ==================================
    Clearly, though, this is very fertile territory for study and there are many interesting things still to be learnt from the rather light nuclei.
    ==================================

    Of course.
    However, first of all they have to solve the primordial puzzle of the nucleon-nucleon force, not solved yet along 66 years.

    After all, how can they solve the puzzles of the light nuclei, since the light nuclei are bound via nucleon-nucleon force, but they did not solve yet the primordial puzzle?

    regards
    wlad

  • Wladimir Guglinski

    REPLY Nr. TWO

    Three-body forces and the binding energy of light nuclei
    http://link.springer.com/article/10.1007%2FBF02710935#page-1
    =========================================
    Summary
    A phenomenological nuclear interaction consisting of a two-body potential containing a Gaussian with exchange and a delta-function, together with a three-body potential of a generalized Gaussian, was fitted to helium 4 and to the single-particle values of oxygen 16. The binding energies of 8Be,12C and 16O were calculated with this potential. The three-body term was found to contribute some five MeV to the binding in oxygen 16. An interaction consisting of a two-body Gaussian with exchange and a three-body Gaussian with exchange was found not to bind oxygen 16
    ==============================================

    Such nuclear model is not able to describe the oxygen 16, because of the following:

    1) The oxygen 17 would be formed by oxygen 16+n

    2) Oxygen 16 has magnetic moment zero, spin zero, and elec. quad. moment zero.

    3) Therefore oxygen 17 would have to have:

    3.1) Magnetic moment not less than -2,10 (-1,913 of the neutron plus 10% due to the rotation of the nucleus). But oxygen 17 has magnetic moment -1,893

    3.2) Spin i=1/2 due to the neutron. However oxygen 17 has spin 5/2

    3.3) Elec. quadr. moment equal to zero, because oxygen 16 has quadrupole moment zero, while the neutron has no charge. However oxygen 17 has quadr. moment equal to -0,26.

    3.4) As it is unable to describe oxygen 17, it makes no sense to suppose that such model can be applied to the oxygen 16.

    CONCLUSION:
    Therefore the model is wrong, and it makes no sense to apply such model to 8Be.

    regards
    wlad

  • Wladimir Guglinski

    Dear Mr. JR,

    I wrote a reply for each one of the links quoted by you, in the total of seven replies.
    I will post them, each one in a different Comment in the JoNP.

    .

    REPLY Nr. ONE

    One of the biggest puzzles in Nuclear Physics starts from the most simplest particle: the deuteron, formed by only a proton and a neutron

    The problem started in 1939, with a paper published in the journal Nature:

    The Electric Quadrupole Moment of the Deuteron and the Field Theory of Nuclear Forces
    http://www.nature.com/nature/journal/v144/n3645/abs/144476a0.html
    —————————————————————-
    |THE discovery by Rabi and his collaborators that the deuteron in its ground state possesses an electric quadrupole moment is of considerable theoretical importance, since it clearly shows that the forces acting between a proton and a neutron must to a quite appreciable extent depend on the spatial orientations of the spins of the heavy particles.
    ——————————————————————-

    .

    In 2005 (therefore 66 years after the publication of the paper by Nature in 1939), the puzzle of the nucleon-nucleon force was not solved yet, as we realize from the paper published by the Tukish Journal of Physics:

    Electron-Deuteron Tensor Polarization and D-State Probability
    http://journals.tubitak.gov.tr/physics/issues/fiz-05-29-3/fiz-29-3-1-0408-3.pdf
    Page 129:
    —————————————————————-
    ”One of the main hopes of electron-deuteron scattering experiments have been to measure certain features of the deuteron wave function and to use these properties to determine unknown properties of the nucleon-nucleon force.
    —————————————————————-

    .

    Well, this is almost unbelievable, because the deuteron is the most simplest bound particle, formed by proton+neutron. After all, after 66 years the theorists were not able to solve the most simplest bound structure existing in the Nature, formed by proton+neutron ?????

    Let us ponder about such unsuccessful attempt made along 66 years.
    If a model is proposed with the same structure existing in the Nature, of course from such model would be viable to describe all the nuclear properties of the structure existing in the Nature.

    However all the attempts along 66 years have failed.
    And a question is unavoidable : WHY ???

    Well, just because the model of forces acting between a proton and a neutron considered in the Standar Model is wrong. If it was right, certainly the problem on nucleon-nucleon force already had been solved.

    So, what happened along 66 years?

    What happened is explained ahead:

    1- A theorist A had proposed the first model
    2- Later other theorists realized that the model of the theorist A was not good, because it was not able to be fit to some results of experiments.
    3- Then a second theorist B had proposed a second model
    4- But later again the other theorists realized that the model of the theorist B was no satisfactory
    5- Then a third theorist C had proposed a third model
    6- And so one…

    And how many models were proposed for the forces acting between a proton and a neutron? Two models? Three? Four? Five?

    In the case of the deuteron, between 1939 and 2005 thirty three models were proposed.

    Thirty three models ?????

    Thirty three models for the most simplest particle formed by proton+neutron ????

    Yes, thirty three models, as we see in the page 129 of the paper published in the Turkish Journal of Physics:
    —————————————————————-
    ”So, in our investigation of deuteron tensor polarization we employ thirty-three local potential models of the nucleon-nucleon force.”
    […]
    “These thirty-three potential models have different deuteron properties, such as deuteron quadrupole moment Q , D-state probability P , asymptotic D-state amplitude A and asymptotic ratio E. The values of these properties are not equal, but have wide range of values in all potential models.”
    […]
    ”To discuss the properties of various wave functions of these potential models, the deuteron radial wave functions u and w of fourteen selected local potential models among the above mentioned 33 potential models (i, c, f, GK3, TSC, r6, RSC, RHC, r7, HJ, PARIS, MHKZ, 2 and 4) are chosen. “
    —————————————————————-

    So, along 66 years the theorists did not succeed to solve the puzzle of the electric quadrupole moment of the deuteron, and the puzzle of the nucleon-nucleon force, and the puzzle is not solved yet.

    As they did not succeed to solve the puzzle of the most simplest nucleon formed by proton+neutron, what can we expect when we consider more complex structures, as for instance the 4Be8 ?

    This is what we will discuss in the next reply to you, Mr. JR.

    Regards
    Wlad

  • Andrea Rossi

    Frank Acland:
    Your question must be divided in different well distinguished points:
    1- Yes, we are studying the paper of Gullstroem
    2- We are studying at 360° all the possibilities of reconciliation between the Standard Model and the results of the ITP Report
    3- I am studying the possible theoretical bases of the so called “Rossi Effect”, that I made through a Galilean methodology of trial and error. As a matter of fact I think a theory is ready, but it is strictly bound to particulars of the reactor that cannot be disclosed so far. I am working upon the issue in collaboration with nuclear physicists.
    4- To create a satisfactory product without a full understanding of the theory behind it is not just possible, it is what happens most of times for most of the inventions: 2226 years ago Archymedes has burnt the fleet of the enemies assailing Syracuse using mirrors that reflected the sunrays: a full understanding of the theory behind this “Archymedes Effect” has been found after circa 2120 years, with the discovery of photons. But the product worked pretty well, ask the enemies of Syracuse!
    Obviously, the same principle is valid for the E-Cat.
    Warm Regards,
    A.R.

  • Frank Acland

    Dear Andrea,

    You have mentioned how you are very interested with the recent Gullstrom paper, and you and your team is studying it carefully. This suggests you are still learning about the theoretical basis of the E-Cat.

    Can you create a satisfactory E-Cat product without a full understanding of the theory behind it?

    Many thanks,

    Frank Acland

  • JR

    Joe,

    I think it’s confusing and somewhat misleading to say that objects have both wave and particle properties (or change back and forth between the two). In quantum mechanics, it’s more correct to say that all objects are always waves, but that the behavior of the waves is similar to particle behavior under many circumstances. This refers to an object’s wavefunctions, not just simple propagating waves.

    As for tunneling, the wavefunction exists on both sides of the barrier; most of it is “inside” the barrier, but a little bit of the wavefunction can cross the barrier and extend beyond. If you measure to see where the object is, there is a small chance (related to the amount of wavefunction beyond the barrier) that you’ll find that it’s on the outside.

    But it’s like Schrodinger’s cat: the object is both inside and outside of the barrier. But then when you look, you’ll either find it inside or outside.

    Of course, this is just a simple description of the quantum mechanics for this problem. Quantum mechanics has many subtleties, and it’s not easy to explain what it means for an object to be a wave or exist both inside and outside of the potential barrier.

  • Joe

    To all the Readers,

    1. Although we know that tunneling is probabilistic, what is the mechanism involved that keeps an object from crossing a barrier some of the time, and allowing it to cross at other times? In other words, why does tunneling not occur at EVERY possible opportunity?

    2. The Schrodinger equation is usually used to describe how a wave tunnels through a barrier. Since an object has both wave and particle properties, how is particle nature transmitted through a barrier? An analogy would be a rubber ball bouncing off a wall, transmitting some of its energy to the other side by way of its wave nature, but having its particle nature prevented from crossing the barrier. It seems that science treats the particle nature as just a toss-in, never really explaining how it makes its appearance on the other side of the barrier. Some people might claim that the particle nature just simply disappears for the time of the crossing, or that the particle nature converts to wave nature for the time of the crossing. But I find these two potential claims illogical. In the case of disappearance, no mechanism exists to even nullify one particle property, let alone many – and that simultaneously. In the case of conversion to wave nature, it is absurd that a confusion of natures would exist. Particle is particle because it is never wave. Even if such conversion did happen, the problem remains that all particle properties would have to convert simultaneously. And no mechanism exists that can accomplish this.

    All the best,
    Joe

  • Steven N. Karels

    Wlad and JR,

    I have enjoyed reading your posts and friendly blanters. JR – thanks for the latest posting with references. I shall read them and attempt to increase my limited (and dated) understanding of the nucleus and Bound vs Unbound effects.

  • JR

    Sorry, Wlad, I never mentioned atoms; you’re the one that thought that they were somehow relevant and introduced them into the discussion (and then immediately complained because they aren’t relevant). However, the binding energy of 4He is absolutely central to the question at hand – if 4He were just slightly less bound, then 8Be would be bound. To assert that the binding energy of 4He is irrelevant whether 8Be is bound or not is to admit to not understanding the question at hand. Of course, claiming that all N=Z even-even nuclei must be bound in conventional nuclear physics is admitting to not understanding much about nuclear physics, which puts you at a disadvantage right out of the gate.

    I could suggest that *you* stop with the nonsense, but of course that’s never going to happen. In any case it’s *far* too late to convince people that you head is functioning properly.

    But to humor you, you can find papers on ab initio structure of nuclei (Variational Monte Carlo, Coupled Cluster, Green’s Function Monte Carlo, Density Functional Theory, etc…), all of which allow you to calculate nuclear structure (with approximations, especially for heavy, complex nuclei). They require input such as nucleon-nucleon potentials extracted from nucleon–nucleon scattering data (chiral, Av18, CD-Bonn, Nijmegan, etc… potentials). Of course, three nucleon forces are important, as I mentioned in my last email, and they aren’t constrained as well (but well enough to describe dozens and dozens of ground state masses and excited state energies with remarkable precision).

    http://www.scholarpedia.org/article/Clusters_in_nuclei gives many references, as does a simple google search: http://bit.ly/1q0r0dB

    A few hastily selected examples follow. Some may not actually include the calculations, but likely they can be found in the references. Others are on the topic of light nuclei in general and may or may not include 8Be. If you want to find specific things, do your own google search, or try actually reading something about the field that you’re constantly bashing with little to no understanding of where the field actually stands today. Even the basics as described on Wikipedia or scholarpedia would be a huge step forward for you if you don’t want to read actual scientific literature.
    http://link.springer.com/article/10.1007%2FBF02710935#page-1
    http://www.cenbg.in2p3.fr/heberge/EcoleJoliotCurie/coursannee/cours/W-Catford.pdf
    http://journals.aps.org/prc/abstract/10.1103/PhysRevC.62.014001
    dx.doi.org/10.1093/ptep/ptu012
    http://www.worldscientific.com/doi/abs/10.1142/S0217732314500278
    dx.doi.org/10.1016/j.nuclphysa.2012.01.010
    http://www.sciencedirect.com/science/article/pii/0375947488901248
    http://link.springer.com/article/10.1140%2Fepja%2Fi2006-10214-6

    Of course, these are complicated; if you want to understand the calculations (and obviously you don’t), it’s best to look at the approach for A=2 and A=3 nuclei. The general input to the calculations is the same for 8Be, but the calculations are much more difficult so approximations and special computational approaches are required in some cases.

  • Wladimir Guglinski

    JR wrote in October 29th, 2014 at 4:43 PM

    Wladimir Guglinski:
    That’s what I thought…

    JR wrote in October 29th, 2014 at 8:43 AM

    So, using your own criterion, I’ll take your refusal to answer as an admission that you don’t know what it means for a nucleus to be bound.
    —————————————–

    That’s what not only I thought… but also what I said…

    Although there is not any mystery with the bound state, of course as a smart politician you are changing the discussion, so that do not show us the THEORY which explains why 4Be8 is unstable, published in a Journal of Physics.

    Bound state
    A nucleus is a bound state of protons and neutrons (nucleons).
    http://en.wikipedia.org/wiki/Bound_state

    As you see, that what you said does not fit to the nucleus:

    By the way, because the 8Be binding is only slightly less than the binding of two 4He nuclei, the calculation has to be very precise to determine if it’s bound or unbound.”.

    Because two atoms of 4He are not a bound state of protons and neutrons, since two atoms of 4He would have to be actually bound by their electrosphere.

    The 4He is a noble gas, the level “s” of the 4He atom is full (2 electrons), and therefore two atoms 2He4 do not form a molecule (with the two atoms sharing the same electron in their orbits), and this is the reason why the binding energy of two 4He atoms is very weak.

    You cannot compare the bound state of two 4He atoms with the bound state of the nucleus 4Be8, since in the 4Be8 the bound state is due to interactions between protons and neutrons.

    Dear Mr. JR,
    I suggest you to stop to say nonsenses, because the reader will start to think that something is wrong with your head.

    regards
    wlad

  • Dear Wladimir Guglinski,
    thank you for the interesting links and suggestions.
    I agree with you that “many nuclear theorists are in the last years realizing that it is impossible to find a theory capable to lead to the nuclear structure, based on the current idea of interactions via strong nuclear force”. However I think the realization is not something recent, and goes back at least a few decades.
    I read the text of Professor Lefteris Kaliambos. I disagree on a series of points, like his proposal of modified quark charges, and the arguments that lead to a negative peripheral charge of the neutron, which is in sharp disagreement with the measurements of Miller in 2007.
    But the analysis that Kaliambos does of the electromagnetic interaction between nucleons and of the progressive assembly of nuclei are interesting.
    I will not enter the debate about the instability of Be8.

    About your second comment/critic. You wrote:
    “If the attractive force was higher than the electrostatic repulsion, then all the isotopes of the all nuclei would have to be stable.”

    The attractive force that Dallacasa proposed depends on:
    – the reciprocal location of the nucleons,
    – the magnetic dipole reciprocal orientation (inclusive precession effects that are not considered in the simplified energy evaluations),
    – the phasing between the rotating charges.

    If the dipoles rotate or the phasing changes, the attraction can decrease and become repulsion. Nuclei are dynamical systems. For the p-p case a simple decrease of the attractive force can lead to the separation of the protons due to the always present long range electrostatic repulsion. In the case of n-n or n-p any positive level of the magnetic attraction leads instead to a bound state. This is the well known reason why progressively heavier nuclei tend to have progressively more neutrons than protons.
    Assembling nucleons in a nucleus means finding the lowest energy combination/s (if any) of:
    – relative average nucleon positions,
    – processing spin direction (magnetic dipole moment),
    – internal rotation phasing.
    Sometimes there are more stable combinations, sometimes there’s only one, sometimes none (e.g. Technetium).

    I have to say that I am always “suspicious” of arguments based on the words “natural” and “normal”. And Prof. Kaliambos uses the expression “NATURAL LAWS”. Anyway it is just my attitude.

    About your first comment. You wrote:
    “The structure of the neutron is n= (p+e), and therefore the lepton number is not violated, since the electron exists into the neutron in the form of a lepton with spin zero (because the electron loses its zitterbewegung into the neutron, and the ZBw is the responsible for the spin 1/2 of a free electron).
    The violation of the lepton number in the Standard Model actually occurs because of the mass of the neutrinos. …”

    Violation of the lepton number conservation would mean that there is something absolutely special in this reaction that differentiates it from the myriad of other cases where the lepton number is conserved. What is it?
    Then you argument a reason for the lack of lepton number conservation. And you say that the electron loses its ZBW inside the neutron. Why? Do you know what causes the ZBW? This would be a major step in physics.
    Let me say that in this case I think you are going against a VERY well proven principle without any VERY well founded explanation. No to mention your claim that the so called Standard Model is WRONG altogether. I would be a bit more cautious in criticizing the result of the interaction of many of the most brilliant minds mankind ever produced.
    Sorry, but I can not consider your arguments. Let me friendly ask you not to try to convince me any further about the non-conservation of the lepton number.

    Best Regards

    Andrea Calaon

  • JR

    Wladimir Guglinski:
    That’s what I thought…

  • Wladimir Guglinski

    JR wrote in October 28th, 2014 at 9:00 PM
    Just for amusement, what do you think “bound” means in this context. Please be specific.

    JR wrote in October 29th, 2014 at 8:43 AM
    Wlad,
    So, using your own criterion, I’ll take your refusal to answer as an admission that you don’t know what it means for a nucleus to be bound.
    ==============================================

    Dear Mr. JR

    I was very specific when I said:

    ——————————————-
    The binding energy of the 4Be8 has nothing to do with the binding energy of two 2He4 nuclei, because:

    1) 4Be8 is a nucleus (the 8 nucleons are packed in one unique nucleus)
    2) two 2He4 nuclei is NOT a nucleus, they are actually two atoms (4 nucleons are packed in one 2He4, and 4 nucleons are packed in other 2He4).
    3) you cannot compare the binding energy of a nucleus with the binding energy of two atoms, BECAUSE ATOMS ARE BOUND VIA ELECTRONS in their electrospheres, AND NUCLEI ARE BOUND VIA STRONG NUCLEAR FORCE (according to the Standard Model).
    ———————————————

    However, Mr. JR,
    as a smart politician,
    after realizing that you said a very stupid thing,
    you tried to deviate the subject of the discussion, by introducing a polemic about the morfology of the word “bound”

    .

    Mr. JR,

    And I repeat again:

    A theory for the explanation on the unstability of the 4Be8 must be proposed as follows:

    1) A nuclear model of the 4Be8 must be proposed
    2) The model must show the distribution of protons and neutrons within the 4Be8, with the energy level of each of them
    3) The model must show the interactions between the nucleons
    4) The unstability of the 4Be8 must be proven from the item 3

    .

    So, Mr. JR
    please show us a nuclear theory where the 4 steps above are fulfilled.

    .

    Unless you come back here and show us a THEORY published in any Journal of Nuclear Physics, I will not waste my time with your claims.

    If you do not to show the THEORY published in a Journal of Physics, I will not waste my time in answering your nonsences.

    regards
    wlad

  • Andrea Rossi

    Frank Acland:
    You bet; in addition I hope our 1MW plant output will give to the Customer of IH the economic profit he has got the plant for.
    Warm Regards,
    A.R.

  • Frank Acland

    Dear Andrea,

    Winter is coming — I hope your R&D output will be enough to cut the heating bill at IH!

    Best wishes,

    Frank

  • JR

    Wlad,

    So, using your own criterion, I’ll take your refusal to answer as an admission that you don’t know what it means for a nucleus to be bound.

  • Andrea Rossi

    Achi:
    He,he,he…nice question.
    It was winter, and in Lugano winter is pretty cold; besides, the laboratory is in a valley between mountains, where cold intensifies. In the photos you cannot see, but along all the ceiling of the laboratory there was a long and big window, that remained open during all the roughly thousand hours of the experiment, so that the hot air mostly escaped through the upper window; nevertheless, the laboratory ( which was pretty big) has been heated enough to force the persons inside to stay in shirts, with an external temperature between minus 5 and plus 10 °C as an average, in the period of February and March. Inside the laboratory the temperature was about of 25°C, but, again, with the hot air , which obviously has a specific gravity minor than the cold air, escaping continuously, 24 hours per day, through the big window of the ceiling of the lab.
    Warm Regards,
    A.R.

  • Wladimir Guglinski

    JR wrote in October 28th, 2014 at 9:00 PM

    Wlad: “The binding energy of the 4Be8 has nothing to do with the binding energy of two 2He4 nuclei”

    So clearly we’re back to Wladimir not understanding the definition of the terms he’s using.

    Just for amusement, what do you think “bound” means in this context. Please be specific.
    ———————————————-

    Dear Mr. JR
    sometimes people choose the wrong profession.

    For instance, I am an engineer, but the best would be if I had chosen to be a physicist.

    While you choose to be a physicist, but the best would be if you had chosen to be a politician. He is an expert in bamboozling people with the power of his speach.

    regards
    wlad

  • Achi

    Dear Mr. Rossi,

    I have a simple question about the test I don’t believe has been answered. I know that you weren’t present during the test much but I think you possibly could have noticed this.
    How was the room ventilated? From the pictures I’ve seen it just looks like a regular room that would get quite hot with the e-cat running 24/7, so i was wondering how they got rid of the heat.

    As you can tell this is mainly just to satisfy my curiosity.

    Thank you,
    Achi

  • JR

    Wlad: “The binding energy of the 4Be8 has nothing to do with the binding energy of two 2He4 nuclei”

    So clearly we’re back to Wladimir not understanding the definition of the terms he’s using.

    Just for amusement, what do you think “bound” means in this context. Please be specific.

  • Andrea Rossi

    Frank Acland:
    Please extend to Carl-Oscar Gullstrom my congratulations for his very intelligent paper. We are going through them with attention.
    If Carl-Oscar Gullstrom contacts me, we can have an exchange of opinion.
    Warm Regards,
    A.R.

  • Wladimir Guglinski

    Silvio,

    Look again what Mr. orene.org says in the link http://forum.rs2theory.org/node/555 :
    ————————————————————-
    “Very interesting theory ring Wladimir Guglinsky considering the Coriolis force would explain how they can interpenetrate two atoms of deuterium to give one of tritium, beating nuclear forces in the cold fusion.”
    ————————————————————-

    While Larson in his book speaks about the “antagonist” force to counter the attractive force, whatever it may be:

    —————————————————–
    “Originally it was assumed that the atoms are impenetrable, and that the electrical forces merely hold them in contact Present-day knowledge of compressibility and other properties of solids has demolished this hypothesis, and it is now evident that there must be what Karl Darrow called an “antagonist,” in the statement quoted in Volume I, to counter the attractive force, whatever it may be, and produce an equilibrium. Physicists have heretofore been unable to find any such force, but the development of the Reciprocal System has now revealed the existence of a powerful and omnipresent force hitherto unknown to science. Here is the missing ingredient in the physical situation, the force that not only explains the cohesion of solid matter, but, as we saw in Volume I, supplies the answers to such seemingly far removed problems as the structure of star clusters and the recession of the galaxies”.
    —————————————-

    In the case of the nuclei, this “antagonist” force can be the Coriolis force, and this is the reason why Mr. orene.org found very interesting the nuclear model proposed in Quantum Ring Theory.

    In the paper Stability of Light Nuclei it is shown that their stabilty is promoted by the the equilibrium between the centripetal force and the magnetic force.
    See for instance the equilibrium of forces in the 3Li6, shown in the Fig. 10 at page 13:
    http://www.journal-of-nuclear-physics.com/files/Stability%20of%20light%20nuclei.pdf

    regards
    wlad

  • Wladimir Guglinski

    JR wrote in October 28th, 2014 at 2:51 PM

    By the way, because the 8Be binding is only slightly less than the binding of two 4He nuclei, the calculation has to be very precise to determine if it’s bound or unbound.
    ———————————————–

    COMMENT

    The binding energy of the 4Be8 has nothing to do with the binding energy of two 2He4 nuclei, because:

    1) 4Be8 is a nucleus (the 8 nucleons are packed in one unique nucleus)
    2) two 2He4 nuclei is NOT a nucleus, they are actually two atoms (4 nucleons are packed in one 2He4, and 4 nucleons are packed in other 2He4).
    3) you cannot compare the binding energy of a nucleus with the binding energy of two atoms, BECAUSE ATOMS ARE BOUND VIA ELECTRONS in their electrospheres, AND NUCLEI ARE BOUND VIA STRONG NUCLEAR FORCE (according to the Standard Model).

    This is one among the most stupid argument used by you up to now.

    .

    The reason why the unstability of the 4Be8 cannot be explained via the Standard Model is very easy to be understood:

    1) All nuclei with Z=N = pair are stable, except 4Be8
    2) In order to explain why only 4Be8 is not stable, there is need to adopt ad hoc hypothesis
    3) However, if we apply the ad hoc hypothesis to the other nuclei with Z=N= pair, like 2He4, 6C12, 8O16, etc, they also cannot be stable, by considering the Standard Model.

    Dear Mr. JR,
    I repeat:

    A theory for the explanation on the unstability of the 4Be8 must be proposed as follows:

    1) A nuclear model of the 4Be8 must be proposed
    2) The model must show the distribution of protons and neutrons within the 4Be8, with the energy level of each of them
    3) The model must show the interactions between the nucleons
    4) The unstability of the 4Be8 must be proven from the item 3

    .

    So, Mr. JR
    please show us a nuclear theory where the 4 steps above are fulfilled.

    And please invite Martin Freer, Nörtershäuser, or any other nuclear theorist to come here to show a theory, based on the 4 steps above.

    .

    Unless you come back here and show us a THEORY published in any Journal of Nuclear Physics, I will not waste my time with your claims.

    regards
    wlad

  • Wladimir Guglinski

    silvio caggia wrote in October 28th, 2014 at 12:22 PM

    @Wladimir Guglinski
    Time ago I asked your question to a RS2 theory guru and seems that this theory predicts that 4Be8 is not stable:
    forum.rs2theory.org/node/555
    Even if I am not able to full understand what he answered me…
    —————————————–

    Silvio,
    interestingly,
    in the link http://forum.rs2theory.org/node/555 mentioned by you there is the following comment on the Quantum Ring Theory, by orene.org:

    ————————————————————-
    “Very interesting theory ring Wladimir Guglinsky considering the Coriolis force would explain how they can interpenetrate two atoms of deuterium to give one of tritium, beating nuclear forces in the cold fusion.”
    ————————————————————-

    regards
    wlad

  • Wladimir Guglinski

    silvio caggia wrote in October 28th, 2014 at 12:22 PM

    @Wladimir Guglinski
    Time ago I asked your question to a RS2 theory guru and seems that this theory predicts that 4Be8 is not stable:
    forum.rs2theory.org/node/555
    Even if I am not able to full understand what he answered me…
    ———————————————-

    Dear Silvio,
    the theory mentioned by the guru is not based on the Standard Nuclear Physics.
    Actually it is a theory proposed by Larson.

    In the page 4 of the book Basic Properties of the Matter, mentioned by the guru, it is written:

    Originally it was assumed that the atoms are impenetrable, and that the electrical forces merely hold them in contact Present-day knowledge of compressibility and other properties of solids has demolished this hypothesis, and it is now evident that there must be what Karl Darrow called an “antagonist,” in the statement quoted in Volume I, to counter the attractive force, whatever it may be, and produce an equilibrium. Physicists have heretofore been unable to find any such force, but the development of the Reciprocal System has now revealed the existence of a powerful and omnipresent force hitherto unknown to science. Here is the missing ingredient in the physical situation, the force that not only explains the cohesion of solid matter, but, as we saw in Volume I, supplies the answers to such seemingly far removed problems as the structure of star clusters and the recession of the galaxies”.
    ———————————————–

    regards
    wlad

  • Andrea Rossi

    Frank Acland:
    Thank you for the information.
    Warm Regards,
    Andrea

  • Frank Acland

    Dear Andrea,

    This article was recently submitted to E-Cat World, and I think you might find it interesting. The title is “Low radiation fusion through bound neutron tunneling”, and is written by Carl-Oscar Gullström, a doctoral student in the Department of Physics and Astronomy at Uppsala University, Sweden.

    https://www.scribd.com/doc/244393652/Low-radiation-fusion-through-bound-neutron-tunneling

    The abstract:

    “To achieve low radiation fusion one considers bound neutron tunneling
    in the MeV range. It is found that the probability for bound neutron
    tunneling is larger then tunneling through a coulomb barrier for Ni Li
    interaction below the energy for fusion conventional Ni Li fusion. The
    theory from basic quantum mechanic tunneling principles are compared
    with the e-cat device. It is found that bound neutron tunneling fusion
    could explain isotope abundance, energy production and burn rate from
    an e-cat test run done by a third party collaboration.”

    Best wishes,

    Frank Acland

  • JR

    Not surprisingly, Wlad once again (a) pretended I was saying something other than what I said and (b) is simply wrong about the state of nuclear theory when he that such models don’t exist.

    First, I was talking about real calculations, not energy differences from mass measurements. Also, one doesn’t obtain the binding energy of 8Be from a direct mass measurement, because it isn’t bound.

    Second, it has been calculated in detailed nuclear models, and all of the things Wlad asks for exist in published papers. That, rather than online comments, is the standard forum for presenting such results. One can search the scientific literature and find several examples, or simply search on “binding light nuclei 8Be” or something similar and find examples (a very quick search found papers from 1998 which provide everything Wlad asked for).

    Wlad’s approach of simply demanding that physicists spend their time correcting his extremely poor understanding of modern nuclear physics in the comment section of various web sites is not very useful. Plus, when nuclear theorists (myself, Martin Freer, Nörtershäuser, etc…) do spend time answering your questions, he just ignores the explanations and arguments that are presented.

    By the way, because the 8Be binding is only slightly less than the binding of two 4He nuclei, the calculation has to be very precise to determine if it’s bound or unbound. So for example, calculations which include only the forces between two-nucleons, and neglect the so-called three-nucleon forces, will not give results that are as precise as one would like. So I’m sure one can search hard enough to find ‘bad’ calculations which give incorrect results, although as a rule, even the less precise calculations still show that 8Be as unbound compared to two alphas.

  • silvio caggia

    @Wladimir Guglinski
    Time ago I asked your question to a RS2 theory guru and seems that this theory predicts that 4Be8 is not stable:
    forum.rs2theory.org/node/555
    Even if I am not able to full understand what he answered me… 🙂

  • Wladimir Guglinski

    JR wrote in October 28th, 2014 at 7:28 AM

    Wlad,

    If you look carefully, I argued that it can be explained in conventional nuclear physics based on the fact that it has been explained in conventional nuclear physics. In particular, by multiple conventional calculations that have shown that it’s unbound.
    —————————————————-

    TO ALL THE NUCLEAR THEORISTS OF THW WORLD:

    There is not any theory for the explanation of the reason why 4Be8 is not stable, because:

    The binding energy of the 4Be8 is obtained as follows:

    1) The mass of the proton is measured by experiments
    2) The mass of the neutron is measured by experiments
    3) The mass of the 4Be8 is measured by experiments

    With the masses measured by experiments, the binding energy of the 4Be8 is calculated.

    THIS IS NOT A THEORY

    It is only an empirical calculation obtained from the results of EXPERIMENTS

    .

    A theory for the explanation on the instability of the 4Be8 must be proposed as follows:

    1) A nuclear model of the 4Be8 must be proposed
    2) The model must show the distribution of protons and neutrons within the 4Be8, with the energy level of each of them
    3) The model must show the interactions between the nucleons
    4) From instability of the 4Be8 must be proven from the item 3

    There is not any theory proposed as required by the 4 steps above

    .

    So,
    I invite all the nuclear theorists of the world, so that to come here to show the theory proposed as shown above.

    .

    regards
    wlad

  • JR

    Wlad,

    If you look carefully, I argued that it can be explained in conventional nuclear physics based on the fact that it has been explained in conventional nuclear physics. In particular, by multiple conventional calculations that have shown that it’s unbound.

  • Andrea Rossi

    Wladimir Guglinski,
    You made your point. Who wants to answer is invited to.
    Warm Regards,
    A.R.

  • Wladimir Guglinski

    JR wrote in October 27th, 2014 at 10:55 PM

    Wlad wrote (again): “It is impossible to explain why 4Be8 is no stable, from the Standard Model”

    I’m sure that this will come as a great surprise to all of the people who have made conventional calculations of the 8Be binding energy and found that it’s not bound.
    ————————————————

    COMMENT

    Dear Andre Rossi,

    in October 20th, 2014 at 10:36 PM you wrote here in the JoNP:

    —————————————–
    Dear Dr Seshavatharam, Dear Prof. Lakshminarayana:
    An answer from you to Wladimir Guglinski appears to be strongly called.
    We’d be delighted to receive it.
    Warm Regards,
    A.R.
    —————————————-

    May you invite them, in order to say to us if the instability of the 4Be8 has explanation by considering the nuclear models of the Standard Nuclear Physis?

    Unfortunatelly our friend Mr. JR does not understand that we cannot use the effect of a phenomenon so that to explain the cause of the phenomenon.

    The small binding energy of the 4Be8 (calculated by using the Einstein’s equation) is consequence of the fact that 4Be8 is not stable.

    As we know, when the Standard Model is no able to explain a phenomenon, Mr. JR uses the inversion of the causality, so that to explain the phenomenon.

    As we know, a man falls ill with ebola when the first ebola virus enter his body.
    But according to Mr. JR the reason why the first ebola virus enter the body of the man is because he was sick with ebola.
    This is the sort of explanation Mr. JR uses in Physics

    So,
    dear Andrea,
    please invite Dr Seshavatharam and Prof. Lakshminarayana, so that to explain to us if we can use the inversion of the causality in the question of the instability of the 4Be8.

    regards
    wlad

  • JR

    Wlad wrote (again): “It is impossible to explain why 4Be8 is no stable, from the Standard Model”

    I’m sure that this will come as a great surprise to all of the people who have made conventional calculations of the 8Be binding energy and found that it’s not bound.

  • Andrea Rossi

    Frank Acland:
    Thank you for your PERMANENT ( hi, Orsobubu) attention.
    One by one, we are resolving all the problems. With patience and dedication, united with the consciousness that if we go through we will have written a page of history.
    Warm Regards,
    A.R.

  • Frank Acland

    Dear Andrea,

    Thank you for your responses. Would you say the problems you encounter are increasing as your work continues, or diminishing?

    Best wishes,

    Frank Acland

  • Andrea Rossi

    Frank Acland:
    Thank you for your continuous attention.
    1- yes
    2- Our team is very strong, all the bases are covered ( electronic engineering, physics, mechanical engineering and top level blue collars). I have good reasons to hope we will not disappoint our Customer, while I am sure we will give the maximum of our skills.
    Warm Regards,
    A.R.

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