A computer program that simulates the structure of atomic nuclei of all known and also possible nuclides.

  • All physics for the study of the atomic nucleus, including CERN, can be compared to firing pebbles at a passing car and studying the range and angles of rebound of these pebbles. Yes, by the way, the car also rotates when moving. In the program, I show the core device as a drawing

  • All physics for the study of the atomic nucleus, including CERN, can be compared to firing pebbles at a passing car and studying the range and angles of rebound of these pebbles. Yes, by the way, the car also rotates when moving. In the program, I show the core device as a drawing

    Can you simulate the full range of electromagnetic bonds possible, including theoretical?

  • I approach the problem of the atomic nucleus from a different angle. Just as any geometric construction can be described by a mathematical formula, so any mathematical formula or theory has a geometric meaning. You're asking about electromagnetic communications... I'm trying to figure out the resulting core design. I see patterns that explain the magic of nuclei and the division into fragments of different masses, the presence and location of clusters. Correlation of experimental data with geometric data. But I don't have anyone to talk to about it. I see an opportunity to get an accurate calculation of the binding energy better than weizsekker and for absolutely all nuclides. Read here -

    https://yadi.sk/i/d6m19-57H6vPQA

    https://yadi.sk/i/vbqRhHEQ63P4mQ


    Official science believes that nuclear forces are very large. This is based on the fact that when a nucleon is incorporated into an atomic nucleus, it simultaneously loses part of its mass. And according to the formula E=mc^2, a very large energy is obtained. I'm waiting for physicists to awaken their knowledge of formal logic. The simultaneity of two physical processes in a limited area of space is certainly a NECESSARY condition for recognizing one process as the cause and the second as the effect. But this is not ENOUGH. To establish a cause-and-effect relationship, you need to know about its mechanism. As far as I know, there Is no such unambiguous understanding. Thus, both of these processes: the occurrence of holding forces and the loss of part of the mass can both be the result of the third, and possibly the fourth... We do not consider a flash of light and an acoustic shock during rain to be a cause and effect. They are both consequences of an electric discharge caused by the ionization of the atmosphere by the solar wind and the earth's magnetic field


    It is the same with nuclear forces. The occurrence of holding forces and the loss of part of the mass of the nucleon may have different reasons. It is quite acceptable to consider them separately. And here the first thought comes about Coulomb forces. Charges distributed in space may well lead to holding forces even for generally neutral structures. Example of a hydrogen atom H2. Just take and calculate the Coulomb forces by taking each Hydrogen atom as three charges. the positive charge of the proton in the center, and two half-negative ones on the radius of their shell of rotation along the interaction line. on the line, rear -1 / 2, center +1, front -1/2, and also for the second Hydrogen. You will be surprised but you will get a curve that is presented in all textbooks as a graph of nuclear forces. And here it is worth considering nucleons as a system with spatially separated charges. And we get the same graph for the Coulomb proton-neutron interaction.


    When you have read both documents, write me your opinion.

  • Nowator ...my suggestion is to re-do your calculations with E = mc to the power n rather using a fixred exponent of 2! This then allows CF to occur. Plug values into Wyttenbach's SO (4) algorithm of reality and there you have all the theoretical answers you need! It's a multidimensional; universe after all!:):):)

  • Взаимодействие с другими людьми

    Nowator ...my suggestion is to re-do your calculations with E = mc to the power n rather using a fixred exponent of 2! This then allows CF to occur. Plug values into Wyttenbach's SO (4) algorithm of reality and there you have all the theoretical answers you need! It's a multidimensional; universe after all!:):):)

    We live in a three-dimensional world. And it turns out that the binding energy values are perfectly correlated with the usual geometric characteristics of the atomic nucleus. And the division into unequal fragments actually occurs by inter-cluster connections. Read the two documents I provided above.

  • An nuclear model which predicts ground state atomic masses is as good as the accuracy of its predictions. Serious models publish standard measures of accuracy (including RMS errors) which are much better than text book graphs. There are plenty of opportunities "to get an accurate calculation of the binding energy [much] better than weizsekker and for absolutely all nuclides".


    See "New global atomic mass formulas", International Journal of Modern Physics B, Vol. 31, No. 25 (2017) for an example claiming an RMS error of 100 keV. This is more than 30 times more accurate than the old Weizsäcker formula of 85 years ago!


    "And it turns out that the binding energy values are perfectly correlated". Perfectly? What are your error figures? Can you provide an algorithm to calculate atomic masses?

  • No, this is where I'm stuck-getting the formula for calculating the binding energy. My thoughts on this issue are in the documents I mentioned above. If you're interested, let's do it together. The documents do not reflect all the patterns I see. But I do not know how to count them. Download the program and see for yourself - download link in the text of the article.

  • o, this is where I'm stuck-getting the formula for calculating the binding energy.

    There is a USA National Nuclear Data Center - NNDC webpage.. maybe you can enquire from them... for their formula..2010.

    http://cdfe.sinp.msu.ru/services/calc_thr/calc_thr.html

    For He 4 Z=2, N=2,

    Result :

    Of course the old concept of 'binding energy' is a bit rough.

    and not very good for elucidating internal structure

    On one Wyttenbach spreadsheet the accuracy of 3.09 eV ..is a bit better than the delta =82500 eV in the "Result"above


    The next Swiss Army knife:) orbital spreadsheet is more exact. (delta= 0.108 eV)

    Tab.9 4He compression with orbits Page 21.

    https://www.researchgate.net/p…context=ProjectUpdatesLog

  • Hi Nowator,


    Can you identify the cause of instability in your model?

    Does your method shed any light on why a nucleus consisting of one, three and four He-4 are stable but two are not? This is a puzzle that I was just wondering about.


    I thought some of your ideas were interesting, like the substructure of nucleons and the reason for reduced mass.


    I have a spreadsheet where I compute what the binding energy of each element would be if it were made up of as many He-4 as possible (contributing only the free He-4 binding energy), then partial He-4 for all the remaining particles including dineutron. The He-4 and partial He-4 account for over 70% of the binding energy per nucleon just using the binding energy for those species when free for all nuclei checked. Can you really have so many unpaired protons and neutrons in your shells? It seems that when you break up a He-4 you lose a lot of binding energy! Here's the sheet: https://drive.google.com/file/…VRm7OxoK/view?usp=sharing


    Do you know of Randell Mills' theory? I have been trying to apply it to nuclei, with modest results. Here is a recent effort https://www.ej-physics.org/ind…ejphysics/article/view/10 and here is Dr. Mills' theory https://brilliantlightpower.com/book-download-and-streaming/ He would agree with you that the nuclear forces are electromagnetic, but has a different geometry for his nucleons.


    Welcome!

  • The nuclear data center publishes only experimental data. There is no mystery. (It's hard for me to explain to a person if You haven't downloaded and opened the simulation program.) about the stability of helium nuclei : Start with Hydrogen. And note that isotopes with an even number of bonds have more energy. Now watch Be8. It has 8 nucleons with three bonds each. Be9-a neutron is added to position Js1 (Shell J position s1) and the distribution of nucleons by the number of bonds changes. Now nucleons with one bond - 1, with three-7 and one with four bonds. And this gives stability to Be9. Similarly with F19, adding a neutron to the Kp1 position increases the number of nucleons with an even number of bonds. And so it is for absolutely all nuclides. I pondered this fact for a long time. Since I am a communications engineer by basic education, I came up with the following mechanism as an assumption : a Nucleon is a three - dimensional cubic resonator with a natural frequency. The frequency value is related to the mass of the nucleon. The phase shift between adjacent faces is 120 or 240 degrees. Between opposites-180 degrees. The oscillation frequencies of the proton and neutron are slightly different, but when attached, they give a resonance at the third frequency located between them. Adding a neutron on the one hand shifts their frequencies to the resonant one. But maksimalnog the mass of each nucleon has its own frequency. A resonant shift in the frequency of the nucleon gives a reduced mass. And each nucleon attached to the nucleus, if it is attached and gives an odd connection, then it reduces the mass by a small value. And if an even bond is attached, the resonance is still shifted and gives a greater reduction in mass. The entire core is a large number of connected resonant circuits with a single frequency. With the addition of each neutron, or a change in its spatial position (for isomers), this frequency changes, determining the total decrease in mass, which is called for some reason the binding energy. Write me your email and I will send You the entire Geometric Theory in what state it is now (requires improvement). I will be glad if you become a co-author.

  • Certainly a very interesting theory of Mr. Dr. J. A. Wittenbach. When I read about toroidal structures, I immediately have a question-what are these structures made of ? There must be the existence of something very small that forms the walls of this torus. Moreover, the small size of these sub-sub-sub particles should form a continuous shape of the walls of these tori. What exactly are the currents in these trades ? What are the microparticles, and what charges to carry these currents and the magnetic field ?

    ---------

    I consider the statement of the difference in the density of different particles to be complete nonsense. I understand that everything about the size of elementary particles depends on the Compton wavelength. There is one animated Soviet film about animals. They measured the length of a boa constrictor by elephants, monkeys, and parrots. As a result, it turned out that in parrots the boa is longer.The Compton wavelength is the wavelength of a gamma quantum with the same Energy as the mass of the particle. That is, we measure a non-relativistic three-dimensional particle (an elephant) by a one-dimensional parrot flying at the speed of light.... Absolute schizophrenia.

  • Quote

    No, this is where I'm stuck-getting the formula for calculating the binding energy.

    Unless we have a generic formula for predicting the atomic mass (or binding energy), we don't have a model at all. A formula which sometimes works on cherry picked cases requires much more work. The late Norman Cook, and more recently Philippe Hatt both claimed geometric models of the nucleus, but required you to know the energies before "calculating" them!! No predictions were possible. If you cannot make predictions, your theory of model is not falsifiable, not scientific.

  • The nuclear data center publishes only experimental data.

    At http://cdfe.sinp.msu.ru/services/calc_thr/calc_thr.html


    The binding energy calculator uses some semiempirical model probably a bit more modern than

    the earlier Weizsäcker models,

    Semi-empirical models are a well trodden path

    Some recent models are


    1995 Duflo Zuker 28 parameter! and 10 parameter models

    discussed at The anatomy of the simplest Duflo-Zuker mass formula

    https://arxiv.org/pdf/0912.0882.pdf

    https://www.researchgate.net/p…13AXJ8KBbp1fbRX-Lg&_iepl=


    2013 EIDIP

    https://www.researchgate.net/p…LEAR_BINDING_ENERGY_MODEL


    2018 Isayev..https://www.researchgate.net/profile/Rafael_Isayev2

    https://www.researchgate.net/p…ION_OF_THE_ATOMIC_NUCLEUS


    in general the fit for helium=4 is much worse than for other isotopes for the semi-empirical models.

    However a good fit for Helium-4 would be a good start to develop a model on..... perhaps a deficit of 100eV?


    The computer zip file does not open on Chromebook.







    ДЕЙТРОННОЙ КЛАСТЕРИЗАЦИИ АТОМНОГО ЯДРА Исаев Р.Ш

    ПОЛУЭМПИРИЧЕСКАЯ ФОРМУЛА ДЛЯ ЭНЕРГИИ СВЯЗИ НА ОСНОВАНИИ ДЕЙТРОННОЙ КЛАСТЕРИЗАЦИИ АТОМНОГО ЯДРА Исаев Р.Ш

  • You can help me. (We will get the Nobel prize together). Tell me how to calculate the centrifugal forces of an object consisting of nucleons located in the nodes of a three-dimensional orthogonal lattice and rotating with it around the z axis. How does the asymmetry of an object affect the centrifugal force ? Whether it causes precession and / or nutation. The nucleons are located in the lattice at a certain distance at the points of equilibrium of Coulomb forces. (we believe that near the neutron - has a small negative charge) How does the displacement of nucleons from equilibrium points affect the binding energy ? How to calculate the centrifugal force acting on the inner nucleon, given that it experiences a centrifugal force and associated nucleons, but located further from the axis of rotation than it. (Analogous to the rotation of a chain of connected balls). I don't know how to count it. All higher mathematics is long forgotten. I am retired and this is my hobby, but it gives interesting results.

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