• Wladimir Guglinski

  October 27th, 2014 at 1:56 PM [www.zpenergy.com]

  Dear Andrea Calaon

  here is other nuclear model in which the equilibrium is via electromagnetism. The author Prof. Lefteris Kaliambos writes:

  “After the discovery of the assumed uncharged neutron (1932) and the invalid relativity (1905) which led to the abandonment of the well-established electromagnetic laws, theoretical physicists developed fallacious nuclear theories for the nuclear force and various nuclear structure models, which cannot lead to the nuclear structure. Under this physics crisis in 2003 I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” by reviving the natural laws which led to my discovery of nine extra charged quarks in proton and nine ones in neutron able to give the nuclear binding and nuclear structure.”
  ———————————————————————-
  http://lefteris-kaliambos.wikia.com/wiki/STRUCTURE_OF_Be8_AND_Be9 [www.zpenergy.com]

  He tries to explain why 4Be8 is not stable by considering his nuclear model working via electromagnetic interactions.

  So,
  it seems 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.

  The reason why 4Be8 is not stable is shown in the item 3.13-5 (Fig. 14 , page 17) of the paper Stability of Light Nuclei:
  http://www.journal-of-nuclear-physics.com/files/Stability%20of%20light%20nuclei.pdf [www.zpenergy.com]

  regards
  wlad

• JR

  October 27th, 2014 at 10:55 PM [www.zpenergy.com]

  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.

• Wladimir Guglinski

  October 28th, 2014 at 5:28 AM [www.zpenergy.com]

  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

  October 28th, 2014 at 7:28 AM [www.zpenergy.com]

  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.

• Wladimir Guglinski

  October 28th, 2014 at 8:56 AM [www.zpenergy.com]

  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 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

  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

  October 28th, 2014 at 2:51 PM [www.zpenergy.com]

  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.

• Wladimir Guglinski

  October 29th, 2014 at 4:05 PM [www.zpenergy.com]

  
  

  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

• JR

  October 29th, 2014 at 7:58 PM [www.zpenergy.com]

  
  

  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 [www.zpenergy.com] gives many references, as does a simple google search: http://bit.ly/1q0r0dB [www.zpenergy.com]

  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 [www.zpenergy.com]
  http://www.cenbg.in2p3.fr/heberge/EcoleJoliotCurie/coursannee/cours/W-Catford.pdf [www.zpenergy.com]
  http://journals.aps.org/prc/abstract/10.1103/PhysRevC.62.014001 [www.zpenergy.com]
  dx.doi.org/10.1093/ptep/ptu012
  http://www.worldscientific.com/doi/abs/10.1142/S0217732314500278 [www.zpenergy.com]
  dx.doi.org/10.1016/j.nuclphysa.2012.01.010
  http://www.sciencedirect.com/science/article/pii/0375947488901248 [www.zpenergy.com]
  http://link.springer.com/article/10.1140%2Fepja%2Fi2006-10214-6 [www.zpenergy.com]

  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

  October 30th, 2014 at 7:39 PM [www.zpenergy.com]

  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 [www.zpenergy.com]
  —————————————————————-
  |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 [www.zpenergy.com]
  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

• Wladimir Guglinski

  October 30th, 2014 at 7:42 PM [www.zpenergy.com]

  REPLY Nr. TWO

  Three-body forces and the binding energy of light nuclei
  http://link.springer.com/article/10.1007%2FBF02710935#page-1 [www.zpenergy.com]
  =========================================
  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

  October 30th, 2014 at 7:45 PM [www.zpenergy.com]

  REPLY Nr. THREE

  The Very Rich Structure of the Rather Light Nuclei
  http://www.cenbg.in2p3.fr/heberge/EcoleJoliotCurie/coursannee/cours/W-Catford.pdf [www.zpenergy.com]
  =======================================
  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

  October 30th, 2014 at 7:49 PM [www.zpenergy.com]

  REPLY Nr. FOUR

  Quantum Monte Carlo calculations of A=8 nuclei
  http://journals.aps.org/prc/abstract/10.1103/PhysRevC.62.014001 [www.zpenergy.com]
  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

  October 30th, 2014 at 7:51 PM [www.zpenergy.com]

  REPLY Nr. FIVE

  Alpha decay constant of 8Be nucleus
  http://www.worldscientific.com/doi/abs/10.1142/S0217732314500278 [www.zpenergy.com]
  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 [www.zpenergy.com]
  ================================================

  
  

  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

  October 30th, 2014 at 7:53 PM [www.zpenergy.com]

  REPLY Nr. SIX

  Energy levels of light nuclei A = 5−10
  http://www.sciencedirect.com/science/article/pii/0375947488901248 [www.zpenergy.com]
  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

  October 30th, 2014 at 7:55 PM [www.zpenergy.com]

  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 [www.zpenergy.com]
  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

  October 31st, 2014 at 8:01 AM [www.zpenergy.com]

  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 has 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, unlike 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 [www.zpenergy.com]
  —————————————————————-
  “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 [www.zpenergy.com]
  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 [www.zpenergy.com]

  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
  

• Wladimir Guglinski

  November 4th, 2014 at 6:44 PM [www.zpenergy.com]

  Why is centripetal force neglected in Standard Nuclear Physics?

  Dears Mr. Joe , Mr. JR , and dr. Stoyan Sarg

  As is known, the centripetal force on the protons and neutrons is not considered in the Standard Nuclear Physics.

  However, a simple calculation shows that centripetal force within the nuclei can have a higher magnitude than Coulomb repulsion . Let us see the calculation.

  Units used:
  Charge of the proton: 1,6×10^-19 C
  Mass of the proton and neutron: 1,7×10^-27 kg
  K= 9×10^9 Nxm²/C²

  I will consider the velocity of the protons 3% of the speed of light c=3×10^8m/s, and so their speed is v= (3×10^-2)x(3×10^8) = 9×10^6m/s.

  Actually the speed of protons due to the rotation of the nuclei cannot be lower than 10% of the light speed, but we will be conservative, and so let consider only 3%.

  Let us consider the nucleus 2He4, by considering the two protons with a distance of 2fm between them (2fm = 2×10^-15m).

  1- Coulomb repulsion between the two protons

  Fe = K.q²/R² = 9×10^9 x (1,6×10^-19)²/(2×10^-15)² = 50N

  2- Centripetal force on each proton

  Fc = 1,7×10^-27 x (9×10^6)²/2×10^-15 = 70N

  .

  So, by considering the speed of protons to be 3% of the light speed (which is an underestimated value), the centripetal force on each proton within the 2He4 has the same magnitude of the Coulomb repulsion force between the two protons.

  If we consider the velocity of protons in the order of 10% of the light speed, we get Fc = 510N (one order of magnitude stronger than the strong nuclear force in a distance of 2fm).

  .

  Obviously the influence of the centripetal force is stronger in other nuclei, as for instance 11Be, where there is a halo neutron moving with radius R=7fm about the cluster. As the strong nuclear force does not actuate in a distance of 7fm, the halo neutron in the 11Be would have to be quickly expelled from the nucleus 11Be, due to the centripetal force on it, since the centripetal force increases with the radius: Fc = m.w².R , where “w” is the angular velocity.
  In the 11Be the centripetal force on the halo neutron is 145N, while the strong nuclear force is practically zero.

  And the situation becomes worst, because the neutron decays in a proton, and the 4Be11 transmutes to 5B10 with a halo proton with orbit radius R=7fm.
  So, beyond the 145N due to centripetal force there is the actuation of a Coulomb force a little weaker than 50N, while the strong nuclear force is practically zero.

  The halo proton would have be expelled quickly from the newborn 5B10, and so 5B10 would have to decay. But this not happens, because the proton actually goes back to the cluster, and the 5B10 becomes stable.

  .

  So, the question is:
  Why the nuclear physicists neglect the centripetal force on the protons and neutrons????

  I hope to hear a good explanation from Mr. JR , or any nuclear theorist he wishes to invite come here to explain it to us.

  Regards
  wlad

• JR

  November 4th, 2014 at 8:38 PM [www.zpenergy.com]

  Wlad said:

  “Dears Mr. Joe , Mr. JR , and dr. Stoyan Sarg

  As is known, the centripetal force on the protons and neutrons is not considered in the Standard Nuclear Physics.”
  ———————–
  Wrong. That was an easy one!

  Of course, even if it were true, almost everything else you said following this was wrong. The 4He nucleus is spin zero so there isn’t a large angular momentum as you’re assuming, a centripetal force CAUSES binding, the centripetal force IS the nuclear force, having the centripetal force get larger than the coulomb doesn’t matter much since the coulomb effect is small compared to the binding from the strong force, etc…

  The closest you come to a true statement is when you *assume* that 11Be can’t be bound by the strong force and then make the bold and daring conclusion that, if there is no binding, then it would not be bound. Bravo.

  I don’t think you could have been more wrong. Actually, you could have been more wrong (and certainly will) simply by saying more. I eagerly await to see what simple ideas you screw up next…

• Daniel De Caluwé

  November 5th, 2014 at 3:53 AM [www.zpenergy.com]

  @JR,

  Wladimir wrote ‘centripetal force’, but I’m sure he meant ‘centrifugal force’, working – in this case – in the same direction as the Coulomb repulsion (of the two protons).

  Kind Regards,

• JR

  November 5th, 2014 at 7:31 AM [www.zpenergy.com]

  Dear Daniel,

  If he meant centrifugal force, then his argument still makes no sense. The centrifugal force is a fictitious force associated with an orbiting body and is caused by the force that pulls the body towards the center of the orbit – the binding force (strong force) in this case. So it fundamentally makes no sense to say that the binding force has to overcome the centrifugal force to maintain a bound system.

• Wladimir Guglinski

  November 5th, 2014 at 1:18 PM [www.zpenergy.com]

  JR wrote in November 4th, 2014 at 8:38 PM

  1) ——————————-
  The 4He nucleus is spin zero so there isn’t a large angular momentum as you’re assuming
  ———————————-

  You are wrong.
  the 4He nucleus has spin zero because one proton has spin up and the other proton has spin down, while one neutron has spin up and the other neutron has spin down. The total spin is zero.

  However all the nuclei have rotation, and so each nucleon (proton or neutron) is submitted to the centripetal force.

  2) ———————————-
  a centripetal force CAUSES binding, the centripetal force IS the nuclear force, having the centripetal force get larger than the coulomb doesn’t matter much since the coulomb effect is small compared to the binding from the strong force, etc…
  ————————————-

  No, according to the Standard Model, which causes binding is the strong nuclear force (and it is not centripetal, i.e., it does not point out to the center of the nucleus, since there is also attraction between two neighbors nucleons).

  3) ———————————-
  The closest you come to a true statement is when you *assume* that 11Be can’t be bound by the strong force and then make the bold and daring conclusion that, if there is no binding, then it would not be bound. Bravo.

  I don’t think you could have been more wrong. Actually, you could have been more wrong (and certainly will) simply by saying more. I eagerly await to see what simple ideas you screw up next…
  ————————————-

  Bravo, Mr. JR, you are using the Heisenberg phantasmagoric method, so that to explain how the cluster of the 11Be can keep a halo neutron without any sort of attraction force between them.

  Heisenber awarded the Nobel Prize with his phantasmagoric method.
  There is a good chance you may get the Nobel Prize too.

  regards
  wlad

• Wladimir Guglinski

  November 5th, 2014 at 2:01 PM [www.zpenergy.com]

  Daniel De Caluwé wrote in November 5th, 2014 at 3:53 AM

  @JR,

  Wladimir wrote ‘centripetal force’, but I’m sure he meant ‘centrifugal force’, working – in this case – in the same direction as the Coulomb repulsion (of the two protons).
  —————————-

  Dear Daniel
  I avoided to call it centrifugal force because the centrifugal force does not exist, I was sure Mr. JR would use it so that to refuse my argument.

  .

  =============================================
  JR wrote in November 5th, 2014 at 7:31 AM

  Dear Daniel,

  If he meant centrifugal force, then his argument still makes no sense. The centrifugal force is a fictitious force associated with an orbiting body and is caused by the force that pulls the body towards the center of the orbit – the binding force (strong force) in this case. So it fundamentally makes no sense to say that the binding force has to overcome the centrifugal force to maintain a bound system.
  ——————————————

  Therefore, according to Mr. Jr,
  when a Formula 1 driver makes a turn too fast, instead of being vented out of the curve, the car is pulled towards the center of curvature … ha ha ha

  My God …
  … then the designers of Formula 1 tracks are crazy, because they put barricades and tires on the outside of the track, to protect against car crashes.

  According to Mr. JR, the designers had to put these barricades on the inside of the runway, so cars do not be thrown into the center of the trajectory …
  ha ha ha

  Dear Mr. JR
  your lack of knowledge of elementary physics is awesome.

  In spite of the centrifugal force is a fictitious force, however due to the rotation of the nucleus the protons and neutrons are submitted to the tendency to be expelled from the nucleus, due to the INERTIA of their motion.

  The protons and neutrons try to continue in a straight TANGENTIAL trajectory, and the strong nuclear force on the protons and neutrons have to avoid they be expelled from the nucleus by such INERTIA

  This tendency due to the INERTIA is vulgarly known as centrifugal force. Within the nuclei the INERTIA is contrary to the strong nuclear force.

  The value of such INERTIA tendency is Fc = m.V²/R, and (as I have shown here) it is at least 10 times of magnitude stronger than the strong nuclear force.

  Therefore, the strong nuclear force cannot avoid the protons and neutrons to be expelled from the nucleus, because the action of the INERTIA on them is 10 times stronger.

  regards
  wlad

• Wladimir Guglinski

  November 5th, 2014 at 4:00 PM [www.zpenergy.com]

  JR wrote in november 4th, 2014 at 8:38 PM

  The 4He nucleus is spin zero so there isn’t a large angular momentum as you’re assuming
  ————————————————————–

  The radii quoted by Bethe due to the rotation of the nucleus as a whole, as if it were a solid body? Be it as it may, numerous measurements had been made by 1936.

  The magnetic moments of nuclei
  Bethe explains that the magnetic moments of numerous are known, thanks to a remark made by Pauli, who showed that if the nucleus behaves as a small magnet, it can somewhat perturb the observed atomic spectral lines caused by electrons making transitions from one quantum state to another. This was called the “hiperfine structure” of the spectral lines, and the magnetic moment of the nucleus could be deduced from it. The magnetic moments of about thirty nuclei were known in 1936.

  page 347
  http://books.google.com.br/books?id=IJa4afSc-MsC&pg=PA347&lpg=PA347&dq=nucleus+rotation+hans+bethe&source=bl&ots=_QY1zPUjBj&sig=XYZNgCuUdsHRJw_biDW6M_mvqEM&hl=en&sa=X&ei=0IVaVPaBGYGUNtaqgPgI&ved=0CEgQ6AEwCA#v=onepage&q=nucleus%20rotation%20hans%20bethe&f=false [www.zpenergy.com]

  So,
  the nuclei have rotation, and since it influences even the atomic spectral lines, the rotation must be very fast, because the magnetic moment of the electron is 9284 (x10^-27 J/T) , while the magnetic moment of the proton is only 14 (x10^-27 J/T).
  Without a large angular momentum the protons and neutrons could not influence the spectral lines (the distance between the electrons in the electrosphere and the protons in the nucleus is 10^-11m , while the size of the proton is 10^-15m, and so the size of the proton is despicable regarding its distance to the electrons). And the magnetic moment decreases with the square of the distance.

  Therefore there are two alternatives:

  1- Mr. JR does not know the nuclear properties of nuclei

  2- Mr. JR knows them, however he tries to deceive people by lying

  regards
  wlad

• Wladimir Guglinski

  November 5th, 2014 at 4:18 PM [www.zpenergy.com]

  Daniel De Caluwé wrote in November 5th, 2014 at 3:53 AM

  @JR,

  Wladimir wrote ‘centripetal force’, but I’m sure he meant ‘centrifugal force’, working – in this case – in the same direction as the Coulomb repulsion (of the two protons).
  —————————————–

  Daniel,
  let me explain it by an easy words.

  Suppose you take a string and you tie a stone with mass “m” in its end.
  And you put the stone moving with circular trajectory with speed V and radius R around your hand.

  The stone tries to escape, applying a force on the string. So, you have to apply a force on the string, otherwise the stone go away with the string.

  The force of the stone is given by Fc = m.V²/R.
  Let us call it centrifugal force (note that, in spite of it is ficticious, however it is able to cause the rupture of a string, because it is actually due to the inertia of the stone, and the inertia is no ficticious).

  The string is able to support a force Fs.

  If the speed of the stone increases so much, the centrifugal force Fc will be stronger than the force Fs of the string, and the string will have a rupture, and the stone will go away.

  Now let us apply it to the nucleus, as follows:

  1- The stone plays the role of a proton

  2- The string plays the role of the strong nuclear force

  The strong nuclear force must be stronger than the centrifugal force Fc , otherwise the proton will move away, leaving the nucleus.

  I showed by calculation that with a speed 10% of the light speed, the centrifugal force Fc on the proton is 500N, while the Coulomb force is 50N.

  In the distance of 2fm (the radius of the nucleus 2He4), the Coulomb repulsion has the same magnitude of the strong nuclear force ( 50N ).

  Therefore we conclude that the centrifugal force on the proton is 10 times stronger than the strong nuclear force.

  regards
  wlad

• Wladimir Guglinski

  November 5th, 2014 at 6:08 PM [www.zpenergy.com]

  On the ficticious centrifugal force

  Dear Daniel De Caluwé

  As the centrifugal force is ficticious, how can it cause the rupture of a string ?

  Let me explain it.

  Suppose you wishes to cause the rupture of a string A with your two hands. So, we have to apply on the string A two contrary forces with your hands, in order to cause its rupture.

  Now let us to do an analogy with the case of a stone moving in circular orbit tied to the end of a string B, while you hold the other end with your hand.

  With analogy to the rupture of the string A with your two hands (where two contrary forces are applied), it seems that two contrary forces must be applied on the ends of the string B. One force is applied by your hand, and the other force is applied by the stone (the centrifugal force, acting in contrary direction of the force applied by your hand).

  Before the rupture of the string B, the force applied by the stone and the force applied by your hand must be equal (since the string was not disrupted).

  But let us analyse it by applying Newton’s law. As the string B is submitted to two contrary and equal forces, the resultant on the string B is zero, and therefore it must be at rest (or to move in rectilinear motion).
  So,
  the centrifugal force does not exist, it is ficticious.

  There is only one force: it is the force applied by your hand. And what is done by this force?
  Well, in each fraction of time such force applied by your hand changes the direction of the stone motion. In other words, you need to apply a force (transmited by the string B to the stone) in order to change every time the direction of the motion of the stone.

  In general, when we have to analyse a phenomenon in which a body has circular motion, the use of the centrifugal force simplifies the analysis and the explanation of the phenomenon. In other words, in spite of we know that the centrifugal force is ficticious, however we use to consider its action, so that to simplify the analysis and the explanation of the phenomenon.

  But sometimes, along a discussion, often we find people like Mr. JR, and they adopt the strategy of refuting our arguments, by claiming that the centrifugal force is ficticious. In this case, there are two situations:

  1- The person uses this sort of argument because he does not understand the discussion

  2- He uses this sort of argument with bad intent, in order to cause confusion to peoples who are reading the debate. By this way, by claiming that the centrifugal force is ficticious and does not exist, he tries to convince the listeners that he is right, and his opposer is wrong.

  So, each reader here has to conclude himself what is the case of our friend Mr. JR.
  I simply wash my hands.

  regards
  wlad

• Daniel De Caluwé

  November 5th, 2014 at 6:28 PM [www.zpenergy.com]

  Dear Wladimir,

  It’s not necessary to explain to me, because I understood you well, from the beginning, when you first calculated the tearing apart of the atom [www.zpenergy.com], due to the centrifugal (and not ‘centripetal’ ) force, caused by the fast motion (around its axis) of the atom, but the only question that remains is this: does the individual atom really rotates so fast??? (Certainly not when it is chemically bond

  Kind Regards,
  Daniel.

• Wladimir Guglinski

  November 5th, 2014 at 7:20 PM [www.zpenergy.com]

  Daniel De Caluwé
  November 5th, 2014 at 6:28 PM

  Dear Wladimir,

  It’s not necessary to explain to me, because I understood you well, from the beginning, when you first calculated the tearing apart of the atom, due to the centrifugal (and not ‘centripetal’ ) force, caused by the fast motion (around its axis) of the atom, but the only question that remains is this: does the individual atom really rotates so fast??? (Certainly not when it is chemically bond
  ——————————————–

  Daniel
  we are not speaking about atoms.
  We are speaking about nuclei.

  The rotation of the nucleus is independent on the rotation of the atom (electrosphere).

  On the Rotation of the Atomic Nucleus
  http://journals.aps.org/pr/abstract/10.1103/PhysRev.53.778 [www.zpenergy.com]

  Wikipedia:
  Therefore there are several possible answers for the nuclear magnetic moment, one for each possible combined l and s state, and the real state of the nucleus is a superposition of them. Thus the real (measured) nuclear magnetic moment is somewhere in between the possible answers.
  http://en.wikipedia.org/wiki/Nuclear_magnetic_moment [www.zpenergy.com]

  A simple model for nuclear rotation at high angular momenta
  Abstract
  A simple solvable model of particles coupled to a rotor is introduced. The solutions illustrate some properties of the nucleus rotating with high angular momentum.
  http://www.sciencedirect.com/science/article/pii/0370269371900578 [www.zpenergy.com]

  Single-Particle and Collective Aspects of Nuclear Rotation
  The spectra of rapidly rotating nuclei reveal two distinct components in the build up of the total angular momentum, corresponding to collective rotation and alignment of orbital angular momentum of individual particles. Various aspects of the interplay of these two mechanisms are discussed. The pattern of collective excitations built upon an yrast state of aligned particle motion is analyzed on the basis of a simple model. For the strongly deformed nuclei, the relative contribution of alignment and collective rotation is characterized by two different moments of inertia referring to the yrast envelope and the collective bands. The behaviour of these moments in the transition region from superfluid to normal phase is considered. Finally, some of the consequences of the build up of angular momentum by alignment and collective rotation are considered for the region of the highest spins, where pair correlations are expected to play a minor role.
  http://iopscience.iop.org/1402-4896/24/1B/001 [www.zpenergy.com]

  Chirality of nuclear rotation
  FIG. 1. The discrete symmetries of the mean field of a rotating triaxial reflection symmetric nucleus (three mirror planes). The axis of rotation (z) is marked by the circular arrow. It coincides with the angular momentum~J.
  http://arxiv.org/pdf/nucl-th/0001038.pdf [www.zpenergy.com]

  regards
  wlad
  

• Daniel De Caluwé

  November 6th, 2014 at 3:41 AM [www.zpenergy.com]

  Dear Wladimir,

  You wrote: Daniel, we are not speaking about atoms. We are speaking about nuclei. The rotation of the nucleus is independent on the rotation of the atom (electrosphere).

  My answer: Yes, I agree, we spoke about the rotation of the nucleus of the atom and not of the atom itself. I realised my mistake after I posted my previous message, but I did not correct it anymore. So my remark about the rotation of the nucleus when the atom is chemically bound (via electrosphere), was a stupid one, because the nucleus still rotates within the atom even when it is chemically bond (via the electrosphere) isn’t it? And as I’m not a nuclear physisist, I did not now that the (or at least some) nucleï rotate so fast. Thank you for the explanation and the interesting links!

  Kind Regards,
  Daniel.

• Wladimir Guglinski

  November 6th, 2014 at 7:27 PM [www.zpenergy.com]

  Daniel De Caluwé wrote in November 6th, 2014 at 3:41 AM

  Dear Wladimir,

  And as I’m not a nuclear physisist, I did not now that the (or at least some) nucleï rotate so fast. Thank you for the explanation and the interesting links!
  —————————————

  Daniel,
  and the situation becomes worst when some nuclei are excited. Their rotation is so fast that it causes the deformation of the nucleus.

  See the figure at left in the title Recent physical results of the article High-spin physics ISP.
  In the figure we see that a spheric nucleus is deformed when it is excited getting a high-spin, and there is a changing in the spheric shape: the nucleus takes the shape of an elipsoid, under the action of the centrifugal force:

  Recent physical results
  First evidence of magnetic rotation in nuclei around mass A = 80
  “The conventional concept of nuclear rotation is based on the existence of a deformed mass distribution of the nucleus (see left figure).”
  https://www.hzdr.de/FWK/MITARB/rs/highspin.html [www.zpenergy.com]

  Therefore,
  under very fast nuclear rotation, the magnitude of the centrifugal force on protons and neutrons is very larger than that of the strong nuclear force (as calculated by me here in the JoNP).
  Note that the nucleus is deformed due to the fast rotation, and therefore the excited high-spin nuclei would have to be desintegrated under the action of the centrifugal force, if protons and neutrons were bound via the strong nuclear force.

  Nevertheless, those excited high-spin nuclei survive, and it means that the strong nuclear force is not the responsible for the nucleus aggregation.

  regards
  wlad

• Andrea Rossi

  November 6th, 2014 at 10:58 AM [www.zpenergy.com]

  To the Readers, request for help:
  Today I have been informed from an Indian nuclear physicist that, during the Manhattan Project, Oppenheimer and Teller expressed the opinion that cold fusion was a possibility. If true, this is important under a historical point of view, but I have not been able to find a reference of this. Is any of our Readers able to inform us about similar reference?
  Warm Regards,
  A.R.

• Felix Rends [www.zpenergy.com]

  November 6th, 2014 at 11:21 AM [www.zpenergy.com]

  “Today I have been informed from an Indian nuclear physicist that, during the Manhattan Project, Oppenheimer and Teller expressed the opinion that cold fusion was a possibility.”

  Dear Andrea Rossi,

  Perhaps the colleague refers to the Oppenheimer Phillips Process:

  http://goo.gl/h01H3z [www.zpenergy.com]
  http://goo.gl/9n5ZNm [www.zpenergy.com]
  http://en.wikipedia.org/wiki/Oppenheimer%E2%80%93Phillips_process [www.zpenergy.com]

  An exothermic stripping reaction (like for example the Oppenheimer-Phillips process), would explain the transmutations taking place and solve the problem with the coulomb barrier.

  Same Days ago I have had the same idea in the LENR Forum:
  http://www.lenr-forum.com/forum/index.php/Thread/442-Is-Calcium-Rossi%E2%80%99s-Secret-Catalyst/?postID=1811#post1811 [www.zpenergy.com]

  Best regards
  Felix Rends

• Andrea Rossi

  November 6th, 2014 at 11:24 AM [www.zpenergy.com]

  Felix Rends:
  Thank you very much, this is very interesting. This could give further evidence that LENR have right of citizenship in the Standard Model system.
  Warm Regards,
  A.R.

• Wladimir Guglinski

  November 7th, 2014 at 2:27 PM [www.zpenergy.com]

  Gell-Mann versus Edward Teller

  Mallove speaking about Gell-Mann:
  From the principles of Quantum Mechanics cold fusion occurrence is impossible to occur, as stated by the Nobel Laureate Murray Gell-Mann at a public forum (lecture at Portland State University in 1998): “It’s a bunch of baloney. Cold fusion is theoretically impossible, and there are no experimental findings that indicate it exists” 3.
  3- E. Mallove, CSICOP: “Science Cops? at War with Cold Fusion, Infinite Energy, V. 4, No. 23, 1999

  E. Teller:
  McKubre was summoned by Edward Teller. “He didn’t think cold fusion was a reality, but said if it were he could account for it with a very small change in the laws of physics.”
  http://archive.wired.com/wired/archive/6.11/coldfusion_pr.html [www.zpenergy.com]

  So,
  two two laureates Nobel Prize with different opinions on cold fusion.
  Why?

  Simple.
  Gell-Mann was sure the cold fusion is impossible because he was sure the Standard Nuclear Physics was developed under unchanging and correct Laws of Physics.

  While Teller knew that something is missing in the Laws of the Standard Model.

  And there is no need to be a genius like Teller to realize that the Standard Model is not the final theory, since there are unsolved puzzle in Nuclear Physics, and some nuclear properties cannot be explained by the Standard Model.

  Teller also knew that by keeping the Laws of the Standard Model would be impossible to explain cold fusion, and that’s why he said that at least a small change in the Laws of Physics is needed.

  However, even a “small” change in the Laws of Physics always represent a big changing in the Physics.

  regards
  wlad

• Wladimir Guglinski

  November 7th, 2014 at 3:01 PM [www.zpenergy.com]

  JR wrote in November 5th, 2014 at 7:31 AM

  Dear Daniel,

  If he meant centrifugal force, then his argument still makes no sense. The centrifugal force is a fictitious force associated with an orbiting body and is caused by the force that pulls the body towards the center of the orbit – the binding force (strong force) in this case. So it fundamentally makes no sense to say that the binding force has to overcome the centrifugal force to maintain a bound system.
  ———————————————

  Dear readers,
  despite Mr. JR has already lost the discussion here,
  we have to take the opportunity,
  so that to show that,
  AGAIN,
  Mr. JR uses the inversion of the causality in his arguments.

  What Mr. JR said is equivalent to say:

  A stone tied to the end of a chord and moving circularly very fast cannot cause the rupture of the chord, because the centrifugal force is a fictitious force associated with an orbiting body and is caused by the force that pulls the body towards the center of the orbit – the force of the chord in this case.

  Therefore,
  according to Mr. JR,
  a chord tied to a stone moving in circular orbit can NEVER be broken.

  According to Mr. JR, a very fine nylon thread for fishing (0.1mm in diameter) can easily keep a mass 100.000kg moving in circular trajectory, because the line will never break, because the line is subjected to the centrifugal force, which is ficticious.

  Those ones who believe that a fine nylon thread will never be break by a mass of 100.000kg moving in circular trajectory tied to the end of the nylon, they can believe in what Mr. JR says here in the JoNP.

  regards
  wlad

That’s why in the Standard Nuclear Physics the electric field of the particles as the proton, the electron, and also the electric fied of the nuclei, is considered as a spherical field involving the particles, or the nucleus.

This electric field of the nuclei considered in the Standard Nuclear Physics is a homogeneous sphere (it means that in any point of the field the value of the electric vector is always the same). Therefore, when a particle as the proton is forced to enter within a nucleus because it is submitted to very high pressure and temperature, the energy necessary to win the Coulomb repulsion is always the same (because no matter where is the point of the electric field of the nucleus where the proton enters, since the energy necessary is the same in any point of the electric field of the nucleus).

Such concept of field adopted in the QFT introduced several puzzles in the Standard Nuclear Physics.
For instance, when an alpha particle (2He4) exits the nucleus 92U238, it leaves out with an energy lower than the energy necessary to put an alpha particle within the nucleus 92U. Such paradox was solved by Gamow. However his solution is not acceptable, because he introduced another paradox in his solution. Besides, if the 2He4 should exit the 92U as proposed by Gamow, because the electric field of the 92U is spherical the 2He4 would have to exit the 92U with a tangential line (because of the rotation of the 92U). But the experiments show that the 2He4 exits the 92U with a radial line.

Other puzzle is the emission of solar neutrinos by the Sun, as we see from the paper published by the journal Nature in 1984:
Solar neutrinos and other problems and their relation to energy production in the Sun:
http://www.nature.com/nature/journal/v312/n5991/abs/312254a0.html [www.zpenergy.com]
“Models of the solar interior, based on the usual physical assumptions, predict a neutrino flux several times greater than that observed in the Davis 37Cl experiment. If, as is widely accepted, this discrepancy represents a ‘flaw’ in the standard solar model one would expect this flaw to manifest itself in other ways also. Here we point out some less well known discrepancies between theoretical predictions of the standard solar model and relevant observations. ”.

The problem is not solved yet, as we realize from the last updated on 23 September 2013 version of the paper The 3He(α,γ)7Be reaction for big bang and stellar nucleosynthesis:
http://www.york.ac.uk/physics/research/nuclear/nuclear-astrophysics/big-bang/ [www.zpenergy.com]

”…; looking for physics beyond standard model of particle physics. Naturally, the reaction attracted early attention of experimentalists and theoreticians alike in the 1950’s. Surprisingly, even today much work needs to be done via nuclear physics experiments to understand this reaction and provide information to the colleagues working on big bang nucleosynthesis, standard solar model and standard model of particle physics”.

Perhaps the puzzle of the rate of solar neutrinos cannot be solved via the Standard Model because they are not considering the cold fusion in the Sun. The emission of neutrinos from cold fusion reactions occurs in a rate very lower than that occurring in hot fusion.

Probably some steps in the nucleosynthesis of some elements in the Sun occurs via cold fusion. And therefore it is impossible to conciliate any theory developed from the Standard Model with the experimental astronomical observations.

.

An experiment published in 2011 proved that space is no empty
The experiment was published in the journal Nature. Light was produced by the space. And therefore the space cannot be empty. It must have a structure, so that to be able to emit light.
A vacuum can yield flashes of light
http://www.nature.com/news/a-vacuum-can-yield-flashes-of-light-1.12430 [www.zpenergy.com]

Therefore the concept of electric field adopted in the Quantum Field Theory must be wrong.

A new concept of electric field, based on the concept of a space having a structure is proposed in the Quantum Ring Theory.
And here a crucial point emerges: there is no way to propose a spherical shape of electric field by considering the space with structure.

Therefore the concept of field adopted in Quantum Field Theory cannot be correct.

Other fundamental puzzle impossible to be solved by considering the concept of field adopted in QFT is concerning the null magnetic moment of even-even nuclei with equal quantity of protons and neutrons, as 2He4, 4Be8, 6C12, 8O16, 10Ne20, etc. Due to the monopolar nature of the electric charge. For instance, the 2He4 has two protons. As the nucleus has rotation, the rotation of the two protons has to induce a magnetic moment. So, by considering the model of field adopted in QFT, it is impossible to explain the null magnetic moment of the 2He4, and all the other even-even nuclei with equal quantity of protons and neutrons.
I had challenged several nuclear theorists for coming to Rossi-Focardi blog Journal of Nuclear Physics-(JoNP) so that to explain how such puzzle could be solved according to the Standard Model. No one nuclear theorist did come.

The reason why the model of field adopted in Quantum Field Theory cannot explain the null magnetic moment of the even-even nuclei with equal quantity of protons and neutrons is because in QFT it is adopted the mono-field model. In my paper Aether Structure for unification between gravity and electromagnetism, submitted for publication in the JoNP, it is shown that the null magnetic moment of those nuclei can also be explained via a double-field model (an outer electric field concentric with an inner central field composed by gravitons), adopted in Quantum Ring Theory.

The Fig. 1 ahead shows the two concentric fields of a proton, as proposed in the paper Aether Structure for unification between gravity and electromagnetism.
FIG. 1
http://peswiki.com/index.php/Image:FIGURE_1-_3_fields_of_the_proton.png [www.zpenergy.com]

As the radius of the electric field has the magnitude of the Bohr’s radius 10^-11m, and the radius of the nucleus is 10^-15m, of course the Fig. 1 does not show the real proportion between the fields. The Fig. 2 show a better proportionality (but of course not real yet):
FIG. 2
http://peswiki.com/index.php/Image:FIGURE_2-_3_fields_in_real_proportionality.png [www.zpenergy.com]

As seen in the Figure 2, there are two “holes” in the electric fields of the particles, and also in the electric fields of the nuclei.
Under suitable condictions of low pressure and temperature, a nucleon as a proton or a deuteron can enter within a nucleus through that hole by having lower energy than that necessary if the nucleon is forced to enter via any other point of the electric field of the nucleus.

By considering that nucleons also may exit a nucleus via the hole in the electric field of the nuclei, we eliminate two paradoxes of the Standard Nuclear Physics:

1) The unacceptable paradox introduced by Gamow, proposed for explaining how a 2He4 can exit the 92U with energy lower than the necessary to cross the Coulomb barrier of the electric field of the 92U

2) Why the 2He4 exits the 92U by a radial trajectory as detected by experiments (impossible to explain from the Gamow theory based on the Standard Model).

.

How a nucleon may enter within a nucleus
Cold fusion may occur by two ways:

1) Via resonance within vessels with conditions of low pressure and temperature, as occurs in the Rossi’s E-Cat.

2) Via kinetic energy in vessels with conditions of very high pressure and temperature, as the Sun. Let us see how it may occur:

a) In the Sun, more than 99,999% of the fusions occur via high nuclear reactions.

b) Cold fusion occurs in less than 0,001% of the nuclear fusions

c) It is very hard to occur cold fusion in the Sun, because the nucleons (for instance a proton) have very high kinetic energy in the star. So, when a proton enters within a nucleus via the “hole” in the electric field of the nucleus, the fusion does not occur (the nucleus cannot capture the proton) because due to the very high kinetic energy the proton simply trespass the nucleus, exiting the nucleus in the other “hole” opposite to the “hole” where the proton had entered.

d) But a cold fusion reaction may occur as follows:

d.1) Within the Sun all the nuclei are moving very fast, and every time changing the direction of their motion due to the collision with other nuclei.

d.2) But suppose that a nucleus (for instance 3Li7) in an exact instant is moving along the x-axis with speed “v”, with the “hole” of its field aligned toward the x-axis. And consider that a deuteron with speed “V” (moving in the same direction along the x-axis) in that exact instant collides against the 3Li7, because the speed V is faster than v. In that condition the relative kinetic energy of the deuteron regarding the 3Li7 is E= 0,5.m.(V² – v²), where “m” is the mass of the deuteron. Therefore the kinetic energy of the deuteron in some very rare conditions is suitable low so that, when the deuteron enters within the 3Li7, the deuteron is captured, and 3Li7 transmutes to 4Be9.

.

The difference between cold fusion and hot fusion
There are three basic differences between hot fusion and cold fusion:

1) Hot fusion occurs when a nucleon enters within a nucleus (only under extreme conditions of high pressure and temperature) when the nucleon succeeds to perforate the electric field of the nucleus.

2) Cold fusion occurs when the particle enters within a nucleus via the “hole” existing in the electric fields of the nuclei. It can occur either in low or in high conditions of temperature and pressure.

3) In order to enter within a nucleus via hot fusion, a particle needs to have a very big kinetic energy. So, when the particle enters within the nucleus, due to the very high kinetic energy of the nucleon the nucleus is excited, and this is the reason why gamma photons are emitted. Unlike, in the case of cold fusion, as the particle enters with low energy the nucleus is not excited, and gamma rays are not emitted. So, also the tax of neutrinos emission in cold fusion is lower than in the case of hot fusion.

Therefore, such property of cold fusion of emitting lower quantity of neutrinos can be response for the question why from the Standard Nuclear Physics there is no way to conciliate the hot fusion reactions in the Sun with the flux of neutrinos emitted by that star.

Conclusion
As we may realize, beyond the challenge of finding a theory capable to explain cold fusion by keeping the principles of the Standard Nuclear Physics there are many other challenges in Nuclear Physics impossible to be solved via the Standard Nuclear Physics.

Up to now the nuclear theorists refused to think about a New Theory based on new fundamental concepts missing in the Standard Nuclear Physics, because of two reasons:
1) Cold fusion is impossible to occur by considering the Standard Model
2) Therefore they were sure it would be possible to solve the unsolved questions by keeping the Standard Model.

But now cold fusion is a reality: Rossi’s Effect was confirmed by three universities of the Europe. And the nuclear theorists worlswide are beginning to accept this new reality, as by one of the top level nuclear phusicist of Russia, Dr Uzikov:
http://www.proatom.ru/modules.php?name=News&file=article&sid=5595 [www.zpenergy.com]

Therefore a reasonable person must realize that it makes no sense to continue trying to explain cold fusion via the old Standard Model. Because the problem is beyond the challenge of finding a theory for explaining cold fusion, actually now the cold fusion became the way for solving the unsolved questions.

Some nuclear theorists have the hope to explain cold fusion via the Standard Model. When we reply to them that cold fusion is impossible to occur from the principles of the Standard Model, they say that we don’t know in deep the Standard Model. However, we may reply to those nuclear theorists: actually you don’t know in deep the structures existing in the Nature. They are very different of that considered in the Standard Nuclear Physics, by beginning from the structure of the space, and as consequence the structure of the electric field of the nuclei, responsible for the difference between hot fusion and cold fusion.