wlad Posted December 7, 2013 Share Posted December 7, 2013 (edited) Heisenberg’s phantasmagoric scientific method is not satisfactory so that to justify why two neutrons do not form a dineutron. Let’s us remember it in short words:a) Two neutrons have a force of attraction due to the strong nuclear forceb) There is not any force of repulsion between two neutronsc) So, two neutrons would have to form the dineutron, because by considering the Classical Nuclear Physics there is not any force of repulsion capable to win the strong nuclear force of attraction between two neutrons.d) Heisenberg proposed the concept of Isospin, so that to justify why the dineutron does not exist. However only a physical force of repulsion would be able to win the physical force of attraction due to the strong force. The Isospin cannot create a physical force of repulsion, because the Isospin is only an abstract mathematical concept.And an abstract mathematical concept cannot separate two neutrons bound by the strong force within the dineutron.The phantasmagoric scientific method inaugurated by Heinsenberg is often used in Physics when a new experiment disproves the current theories, and the theorists cannot find a satisfactory explanation so that to justify why.So, the theorists use the Heisenberg’s phantasmagoric everytime they cannot discover the physical cause responsible for a phenomenon which disproves the current models.Here we will see how Heinseberg’s phantasmagoric method has been used in Classical Nuclear Physics, after the publication of two experiments, one published in 2011, and the other in 2013.1) Pear shaped nuclei =============================According to the current nuclear models, the even-even nuclei have to have two sort of shapes:A) Spherical shape – when the quantity of prótons Z is the same of the quantity of neutrons N, Z=N.B) Elispoidal shape – when Z << N .These two sort of shapes (spherical or ellipsoidal) have to occur for eve-even nuclei in Classical Nuclear Physics because there is not any physical cause which we could find in the current nuclear models, in order to justify any other sort of shape different of the spherical or the ellipsoidal.But experiments have shown that some even-even nuclei with Z<< N are pear shaped.Then, how to justify it? After all, they would have to have an ellipsoidal shape.In 2013 the Professor Peter Butler of the University of Liverpool had proposed that nucleons (prótons and neutrons) are distributed within the nuclei around a z-axis. So, in order to justify why some even-even with Z<<N (as for instance 88Ra224) have pear shape, he proposed the existence of a z-axis within the nuclei.However there is no way to justify why prótons and neutrons are distributed about a z-axis within the even-even nuclei, by considering the current nuclear models.And the reason is obvious: there is no way to find a physical cause responsible for putting the prótons and neutrons distributed along a z-axis in the current nuclear models.Therefore, the Professor Peter Butler had actually used the Heisenberg’s phantasmagoric scientific method, so that to justify why 88Ra224 is pear shaped, by using a phantasmagoric hypothesis: the existence of a z-axis, which existence is impossible to explain by considering the current nuclear models.Look the z-axis proposed by Professor Butler for the 88Ra224 in the link:Scientists demonstrate pear shaped atomic nuclei - University of Liverpool NewsThe distribution of prótons and neutrons about a z-axis had been predicted in my book Quantum Ring Theory, published in 2006 (therefore 7 years before the proposal by Professor Peter Butler).The existence of the z-axis is perfectly justified in the nuclear model proposed in my Quantum Ring Theory because there is a physical cause which obliges the prótons and neutrons to take a distribution about the z-axis: in the nuclear model proposed in QRT there is a central 2He4 which captures the prótons and neutrons, in order that they take a distribution about the z-axis.That’s why the existence of the z-axis had been correctly predicted in my Quantum Ring Theory.See page 198 of my book Quantum Ring Theory: ======================================================.2) Non-spherical even-even nuclei with Z=N ====================Along more than 60 years the nuclear theorists had believed that even-even nuclei with Z=N have spherical shape, because from the current nuclear models we have to expect that they have to have spherical shape, because there is not any physical cause from which a nuclear theorists could justify a non-spherical shape for those nuclei.For instance, in 2009 the nuclear theorist Dr. Martin Freer proposed the spherical shape for the oxygen 8O16 shown in the Figure 1 of his paper Clusters in nuclei published in Scholarpedia:Clusters in nuclei - ScholarpediaBut in 2011 new experiments had detected that even-even light nuclei with Z=N have non-spherical shape.So, Dr. Martin Freer and all the nuclear theorists had to change their mind, after more than 60 years of the nuclear theorists believing that even-even nuclei with Z=N have spherical shape.That’s why in 2012 the journal Nature published the paper How atomic nuclei cluster, where the authors consider several even-even nuclei with Z=N with non-spherical shapes.http://www.nature.com/nature/journal/v487/n7407/full/nature11246.htmlHowever, a question appears:From the current nuclear models we have to expect that even-even nuclei with Z=N have to have spherical shape. Then how does to justify that they have non-spherical shape ??????In other words: what should be the physical cause, according to the current nuclear models, responsible for the non-spherical shape of those nuclei?Well, of course there is no way to find such physical cause from the current nuclear models. And the authors of the paper How atomic nuclei cluster cannot give any physical cause capable to explain the non-spherical shape of those nuclei.Therefore, actually Dr. Martin Freer and the authors of the paper published in the journal Nature are using the phantasmagoric scientific method proposed by Heisenberg, because it is the unique way so that to justify the non-spherical shape of the even-even nuclei with Z=N.The non-spherical shape of the even-even nuclei with Z=N was predicted in my book Quantum Ring Theory (published in 2006, and therefore 6 years before the publication in the journal Nature.The non-spherical shape of those nuclei had been predicted and proposed in my book because there is a physical cause which oblige the even-even nuclei with Z=N to take that shape: the existence of a central 2He4 in all the nuclei.It is opportune to remember that the journal Nature published a plagiarism of an argument proposed in the page 137 of my book Quantum Ring Theory, as I explain again here:A) Even-even nuclei with Z=N with non-spherical shape cannot have null electric quadrupole moment.B) But the experiments detected that even-even nuclei with Z=N have null electric quadrupole moment.C) So, how to justify that even-even nuclei with Z=N have non-spherical shape, in spite of they have null quadrupole moment ?D) In the page 137 of my book I had proposed the explanation: as those nuclei have nuclear spin zero and magnetic moment zero, they gyrate chaotically, and there is no way to align them along an external magnetic Field, so that to measure their quadrupole moment. Therefore, in spite of their quadrupole moment is not zero, however it is not possible to measure it.E) I sent an email to Dr. Martin Freer, telling him that as the journal Nature published a paper where the even-even nuclei with Z=N have non-spherical shape, they would have to exhibit non-null quadrupole moment. And I asked to him to explain why the experiments detect null quadrupole moment.F) Dr. Martin Freer sent me a reply giving the same explanation proposed by me in the page 137 of the book Quantum Ring Theory.Look at the explanation in the page 137 of the book QRT (attached bellow). And the explanation by Dr. Martin Freer:The nucleus is intrinsically deformed as shown, but has spin 0. Consequently, there is no preferred orientation in the laboratory frame and thus the experimental quadrupole is an average over all orientations and hence is zero. Experimentally is is possible to show that the deformation of the ground state is non zero by breaking the symmetry and rotating the nucleus.Martin===================================================3) CONCLUSION =====================================When new experiments defy the current theories, the theorists use the phantasmagoric scientific method proposed by Heisenberg because it is impossible to find the physical causes responsible for the phenomenon detected in the experiment.Of course it is easier to use such phantasmagoric method, so that to adapt the old current models in the results of new experiments which disprove those current models.By such a phantasmagoric method the theorists simply use some abstract mathematic concepts, neglecting the missing of physical causes which are the actual causes responsible for the phenomenon.And so obviously the phantasmagoric method hides some physical cause responsible for the phenomenon detected in the new experiment.In other words: the true cause responsible for the phenomenon is missing in the current models used so that to explain the phenomenon.And the crucial question is the following:Is it possible to find a definitive theory by starting up from such phantasmagoric scientific method proposed by Heisenberg ?Suppose that there is indeed a central 2He4 within the nuclei (as are suggesting the experiments published in 2012 by Nature and in 2013 by Professor Peter Butler).Then let’s do the question:Will the nuclear theorists succeed in their enterprise trying to explain all the nuclear properties, by using nuclear models where it is missing the central 2He4 ? Edited December 7, 2013 by wlad Link to comment Share on other sites More sharing options...
Sensei Posted December 7, 2013 Share Posted December 7, 2013 (edited) Free neutron is unstable particle. It will decay to proton, electron and neutrino within 15 minutes or so. Even if two neutrons could theoretically join, one or both of them, will decay, and you won't have "dineutron" anymore. Edited December 7, 2013 by Sensei 1 Link to comment Share on other sites More sharing options...
wlad Posted December 8, 2013 Author Share Posted December 8, 2013 (edited) Free neutron is unstable particle. It will decay to proton, electron and neutrino within 15 minutes or so. Even if two neutrons could theoretically join, one or both of them, will decay, and you won't have "dineutron" anymore. you are wrong because: 1- two neutrons bound by the strong force would not decay, as happens with the stable deuteron formed by proton-neutron bound by the strong force 2- even if the dineutron would decay in 15 minutes, its existence could be detected by experiments along the 15 minutes of its existence. However the dineutron had never been detected by experiments Free neutron is unstable particle. It will decay to proton, electron and neutrino within 15 minutes or so. Even if two neutrons could theoretically join, one or both of them, will decay, and you won't have "dineutron" anymore. besides, if you had be right, then Heisenberg would not need to propose the concept of isospin. He could use the same argument used by you, saying that dineutron do not exist because one of the neutrons decays. I hope you are not suggesting that Heisenberg was a stupid guy, since he did not propose such argument used by you. Edited December 8, 2013 by wlad Link to comment Share on other sites More sharing options...
Sensei Posted December 8, 2013 Share Posted December 8, 2013 In quark model neutron is made of 2 unstable down quarks, and 1 stable up quark. Your "dineutron" would be made of 4 unstable particles and 2 stable.. Check out how unstable is Delta baryon which has 3 unstable down quarks, and none stable: http://en.wikipedia.org/wiki/Delta_baryon Link to comment Share on other sites More sharing options...
swansont Posted December 9, 2013 Share Posted December 9, 2013 Heisenberg’s phantasmagoric scientific method is not satisfactory so that to justify why two neutrons do not form a dineutron. Let’s us remember it in short words: a) Two neutrons have a force of attraction due to the strong nuclear force b) There is not any force of repulsion between two neutrons The strong force is not so simple as you imply here. There is an isospin dependence, leading to a repulsive potential for either a diproton or dineutron. Link to comment Share on other sites More sharing options...
wlad Posted December 9, 2013 Author Share Posted December 9, 2013 The strong force is not so simple as you imply here. There is an isospin dependence, leading to a repulsive potential for either a diproton or dineutron. repulsive potential ?????? repulsive potential makes no sense, because the neutron has not charge. Obviously as the question of the dineutron is not explained by considering the current Nuclear Physics, the nuclear theorists use some subterfuges, as to say that the strong force is not so simple. But the question is very simple: there is not Coulomb repulsion between two neutrons, and therefore they would have to be bound and form the dineutron. In quark model neutron is made of 2 unstable down quarks, and 1 stable up quark. Your "dineutron" would be made of 4 unstable particles and 2 stable.. Check out how unstable is Delta baryon which has 3 unstable down quarks, and none stable: http://en.wikipedia.org/wiki/Delta_baryon I dont need to check nothing there is not repulsion between two neutrons, and therefore they would have to be bound by the strong force. As always happens when a theory does not fit to experimental findings, the theorists use to propose ad hoc hypothesis which do not fit to the experimental findings. Besides, there is not such a thing as stable up quark. Quarks are not stable. They are stable only when they are confined into the proton. -2 Link to comment Share on other sites More sharing options...
Endy0816 Posted December 10, 2013 Share Posted December 10, 2013 The force is powerfully attractive between nucleons at distances of about 1 femtometer (fm) between their centers, but rapidly decreases to insignificance at distances beyond about 2.5 fm. At very short distances less than 0.7 fm, it becomes repulsive, and is responsible for the physical size of nuclei, since the nucleons can come no closer than the force allows. http://en.wikipedia.org/wiki/Nuclear_force Your argument is bunk. 1 Link to comment Share on other sites More sharing options...
wlad Posted December 10, 2013 Author Share Posted December 10, 2013 The strong force is not so simple as you imply here. There is an isospin dependence, leading to a repulsive potential for either a diproton or dineutron. oh... yes... because the experiments had detected the repulsive potential. In another words: it is an experimental fact. However, there is no way to explain it by considering the current nuclear models, just because the current nuclear models were built from phantasmagoric assumptions, where physical causes are missing. You are making confusion between two different things: 1- One thing is the experimental result 2- Other thing is to explain it by considering the current nuclear models. It is just here where the theory fails. Dong make confusion between experimental fact and the theory used so that to explain the experimental fact. To claim that "The strong force is not so simple as you imply here" is not an acceptable argument solution. Because by considering the current nuyclear models the strong force would have to be so simple as I imply here. In order to explain why the strong force is not so simple there is need to consider physical causes missing in current nuclear models. -1 Link to comment Share on other sites More sharing options...
John Cuthber Posted December 10, 2013 Share Posted December 10, 2013 repulsive potential ?????? repulsive potential makes no sense, because the neutron has not charge. I dont need to check nothing Quite. Link to comment Share on other sites More sharing options...
swansont Posted December 10, 2013 Share Posted December 10, 2013 repulsive potential ?????? repulsive potential makes no sense, because the neutron has not charge. Obviously as the question of the dineutron is not explained by considering the current Nuclear Physics Or, your understanding of basic nuclear physics is in need of some improvement. Some people, when given the opportunity to learn something they didn't know, are happy for the knowledge. Others get angry at being contradicted. The latter generally do not become scientists. Link to comment Share on other sites More sharing options...
John Cuthber Posted December 10, 2013 Share Posted December 10, 2013 Some people, when given the opportunity to learn something they didn't know, are happy for the knowledge. Others get angry at being contradicted. The latter generally do not become scientists. They tend to have a pretty short half life on discussion fora too. Link to comment Share on other sites More sharing options...
wlad Posted December 12, 2013 Author Share Posted December 12, 2013 Or, your understanding of basic nuclear physics is in need of some improvement. Some people, when given the opportunity to learn something they didn't know, are happy for the knowledge. Others get angry at being contradicted. The latter generally do not become scientists. Dear swansont, 1- Repulsive force between two neutrons occurs only in very short distances (about 0,1 fm) 2- There is not repulsion between two neutrons in a distance of 2fm (which is the distance of the strong force actuation) 3- So, two neutrons would have to bound in a distance of 2fm. 4- Even if two neutrons aproach in a distance of 0,1fm, and they have repulsion, however they will move one away of the other, and their distance would increase. With the growth of their distance, the repulsion between them dispapears when their distance becomes 2fm. And in such distance of 2fm they have to be bound by the strong force. What I know about scientists is that generally the most of them use to betray the scientific method. For instance, as when they refuse to accept the Don Borghi and the Conte-Pieralice experiments, both them proving that the neutron is formed by proton+electron. C. Borghi, C. Giori, A.A. Dall’Ollio, Experimental Evidence of Emission of Neutrons from Cold Hydrogen Plasma, American Institute of Physics (Phys. At. Nucl.), vol 56, no 7, 1993. Probably you had never heard about the two experiments. Just because, as the scientists use to betray the scientific by refusing to accept experiments which disprove the current theories, they undertake all the effort so that to hide the existence of those two experiments. regards wlad -1 Link to comment Share on other sites More sharing options...
swansont Posted December 12, 2013 Share Posted December 12, 2013 The interaction is isospin-dependent. You can't just ignore that. Link to comment Share on other sites More sharing options...
wlad Posted December 12, 2013 Author Share Posted December 12, 2013 The interaction is isospin-dependent. You can't just ignore that. Isospin is an abstract mathematical concept. Interaction is a physical phenomenon, and so it cannot be dependent of an abstract concept. That's why Heisenberg scientific method is phantasmagoric. And that's why the experiments made between 2010 and 2013 are proving that current nuclear models are wrong, because the Standard Nuclear Physics had been developed via the Heisenberg's phantasmagoric scientific method. -1 Link to comment Share on other sites More sharing options...
swansont Posted December 12, 2013 Share Posted December 12, 2013 Isospin is an abstract mathematical concept. Interaction is a physical phenomenon, and so it cannot be dependent of an abstract concept. Can't be dependent on abstract concepts? Seriously? I thought we were talking about physics here. You know, the branch of science that discusses (and is based upon) abstract concepts of energy and momentum, time, electric and magnetic fields, spacetime, and more. We even have angular momentum of objects that don't actually physically rotate. Physics. Isospin and the sign difference of the isospin coupling interaction are discussed on p 118 of my copy of Elementary Particles by Griffiths. It's not like this is some obscure part of nuclear physics you would miss if you were actually conversant in nuclear physics. I know about it and it's not even my field! So your objections are pretty much moot, as far as I'm concerned. Link to comment Share on other sites More sharing options...
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