I am seeking collaboration for further developments in approaching the proposed model of the deuteron structure within a quantum mechanical framework, and aiming to form a research group. Those interested please contact me at: [email protected]

The distribution of the structural charge density of the deuteron is examined and its symmetry is analyzed on the basis of experimental data obtained from electron scattering experiments, in which the total symmetry of the deuteron structure is evidenced. Since according to the conventional nuclear model the deuteron is formed of a neutron and a proton, whose structural charge density is substantially different, it follows that the juxtaposition of the distribution of their charge density is asymmetrical, thus being in deep disagreement with that of the deuteron which is symmetrical. This incongruence is thus analyzed. The conventional model of a proton juxtaposed to a neutron is unable to provide a credible explanation of the symmetry of the deuteron charge distribution since it is composed of two different particles, one neutral and the other one charged, and with a highly dissimilar structural charge density. Consequently, an explanation for the structural symmetry of deuteron is proposed, based on a revised approach.

https://www.researchgate.net/publication/330468167_About_the_symmetry_of_the_deuteron_structural_charge_density_distribution

Dear Georges Sardin , you appear to be a very learned person and I have gone through your posts and profile. Your simple language suggests that you might be generous towards daring science enthusiasts like me. I am going to give you a very arrogant reply of what you have asked.

First of all, from a basic semantic sense, let's ask what is symmetry? Symmetry can be seen in many examples, along with real world distortions. For example, wings of a butterfly are highly symmetrical in appearance. In nature, we find many examples of symmetry. If you throw a pebble in a lake, then concentric ripples of radial waves emerge, and they are all radially symmetric figures emerging and disappearing and reemerging again. In general sense, this is symmetry. Now, let's talk about something like charge symmetry. For instance, if an atom is electrically neutral, then we can say that it has equal numbers of +ve and -ve charges. If we disregard any Coulomb forces, then that can be defined as a symmetric situation of charge quantities. This is not the case with mass. Negatively charged electrons have far lesser mass than positively charged proton. Why?

I am not entering the multidimensional space-time situation of sub atomic orbitals and electron clouds at this point. Please be careful and follow the logic I am putting rather than experimental findings. What is the essence of symmetry? Does that apply to structures only? No. Suppose, we are jumbling up a pack of several cards. There are two identical cards for every design in that pack. Now we arbitrarily distribute these cards among two sets. And alas ... suppose we find that each set contains one card for one unique design each! This is difficult, but if this is possible, then we have just managed to distribute the cards into two symmetric sets!

And then there is the simple but staggering magic ... If we wish to create symmetric distributions, then we must have two sets of identical things. For instance, if you reflect yourself on a mirror, then you have two symmetrically opposite images across that mirror. One is the real you, and the other is your reflected image.

And this is the place where we must stumble upon. Complementary configurations, if any, can be found out only inside a set of even number of events, tests, or objects. In other words, there is a requirement of double, and not single. But what about our hydrogen atom? It has only one proton inside it, no neutron, In the case it has a neutron, then its mass is 2 rather than 1 ... and this is why the concept of atomic number (which is crucial in the construction of periodic law) needs to be reviewed.

Generally, atomic number of an element is thought of as the number of protons present in the nucleus of that element. However, for a while, let's redefine stable isotopes of elements as the only ones that have at least one neutron in their respective nuclei. In that case, Hydrogen-1 would no more qualify as an ideally stable element ... that place will go to the isotope Hydrogen-2. In other words, not the 'protium nucleus' but the 'deuteron' should have the status of being the nucleus of a stable element. This kind of reassignment of numerical semantics has the capability of triggering off a whole new periodic law, which would necessitate having at least one neutron in the elemental nucleus. And at that point, Hydrogen would be understood as a 'point zero' and not as the 'starting point.'

I have attempted to write down some paragraphs covering this issue in my paper at Research Speculation of a Modified Exponential Series, based on the M...

Another dangerous possibility is that maybe deuterium (and not protium) is the starting point of the series of elements that we see in our universe today. Possibly that is why it has a symmetric configuration of charged nucleons ... Who knows? After all, asymmetry would start surfacing after a point is reached where symmetry is complete (or structural entropy is zero). Above that point, we have post-symmetric elemental atoms. And below that same point, we are left with the only pre-symmetric elemental atom of Hydrogen-1. Is not this kind of speculation a little bit thrilling? Where are the other pre-symmetric nuclides then?

Dear Arghya,

It looks like you are young and very enthusiastic and that is very nice, but let me stress a few points about your comments:

You said: “If we disregard any Coulomb forces, then that can be defined as a symmetric situation of charge quantities. This is not the case with mass. Negatively charged electrons have far lesser mass than positively charged proton. Why?”

But the symmetric particle of the proton is the anti-proton (or negative proton) which has the same mass as the proton.

“Generally, atomic number of an element is thought of as the number of protons present in the nucleus of that element”.

But they are other standard classifications, as e.g. the isotones (classification according to the neutron content) and isobars (according to the mass number), and still one classify them according to other diverse criteria.

“However, for a while, let's redefine stable isotopes of elements as the only ones that have at least one neutron in their respective nuclei”.

As I have mentioned above, isotones are classified according to the number of neutrons

“Another dangerous possibility is that maybe deuterium (and not protium) is the starting point of the series of elements that we see in our universe today”

But the Sun is almost totally made of hydrogen, as well as stars, and so is a major component of the Universe, so, why should we choose the deuterium as the starting point?

Your enthusiasm for science should not make you too speculative. Physics should be treated as a very rigorous science in which precarious standpoints are soon or later discarded by experiments. But, keep your enthusiasm and your imagination, they are both very valuable, still at the same time, in Science the imagination should be disciplined.

Dear @Georges Sardin , you said that "Your enthusiasm for science should not make you too speculative" ... Well, the problem with me is that I am not a qualified person in scientific field. Of course, when someone goes to a Physics class, he/she has a better knowledge and encounters several eye-openers by professors and guides. General public like me, who did never go to a science college, are mostly speculative (I personally feel so, may be I am wrong). Yet, I will dare to make this point once again.

Hydrogen-1 is highly abundant in our universe. But Hydrogen-2 does not only show isotopic stability, but also has got neutron content. The single factor of 'Protium's abundance over Deuterium' cannot be given 100% priority while evaluating the different mathematical properties associated with the physical exhibitions of the isotopes or elements, whatever may the case be. Maybe I not getting the proper language, but what I wish to say is that mathematical properties, such as orientations or symmetries, might be very much in relationship with possible positions and translations of the the sub atomic particles. In the case of Deuterium, this attribute must be analyzed ... I feel symmetric configurations may lead to stability, and stability is important to be had with the Deuterium atom. But why? Why not Tritium is showing that kind of symmetric manifestations, or why do we not have an isotope like Hydrogen-4?

Please have a look at Abe et al's (2018) work, and also please read the Physics Stack Exchange question by Vaicu (2017) (at the stack exchange too, the administrators have closed that discussion since they also feel that the discussion has not been put clearly).

References

Abe, K., Yamaoka, S., & Hyeon-Deuk, K. (2018). Isotopic Effects on Intermolecular and Intramolecular Structure and Dynamics in Hydrogen, Deuterium, and Tritium Liquids: Normal Liquid and Weakly and Strongly Cooled Liquids. The Journal of Physical Chemistry B, 122(34), 8233-824

Vaicu, S. (2017), Strange behavior difference between light and heavy hydrogen atoms [closed]. Retrieved on 15th February 2019 from https://physics.stackexchange.com/questions/369679/strange-behavior-difference-between-light-and-heavy-hydrogen-atoms

Dear Arghya Ray,

For someone who did never go to a science college you are doing it pretty well. I congratulate you. It looks like you have a very sharp mind. Thanks for the references and the link you pointed me. I had a look to the link but when I will find some more time I will read them carefully and think over the related issues.

I just mention you that it has been identified 7 isotopes of H, but of course the heavier ones have a very short life-time. A very useful site for physics data is that of CODATA (NIST): https://physics.nist.gov/cuu/Constants/

Georges Sardin ... You are great sir, you have understood my tendency very well. "I just mention you that it has been identified 7 isotopes of H, but of course the heavier ones have a very short life-time." I did not know about this. I will see them when I get time. I know almost nothing about recent developments in theoretical research across the physical sciences and pure mathematics. You have inspired me so very much, so I will dare to make another comment or suggestion. Please, if possible, make use of Transformation Geometry to re-analyse the theoretical context of quantum mechanics and atomic structures. I believe that before delving into more critical issues like super-symmetry, we would need to implement at least some concepts from transformation geometry directly onto quantum mechanics. But I do not have so much knowledge so as to do this. Please, if you have some spare time, think about it. For more information on transformation geometry, please visit https://en.wikipedia.org/wiki/Transformation_geometry. Hope learned people like you would definitely forgive the arrogance among enthusiasts like me.

I would suggest that the deuteron is not composed of two different-type particles. It is a femto-molecular ion consisting of two protons sharing (and therefore bound by) a deep-orbit highly relativistic electron. Thus, averaged over time (even extremely brief periods), it is symmetric.

The problem arises from the fact that the electron has a mass and therefore a relativistic electron spinning around two protons would have an enormous centrifugal force, much higher than any cohesive force between them. Hence the electron cannot act as a cohesive element and thus the two protons could not be bound.

The electrostatic force is not appropriate. You need the electro-dynamic force, which is enhanced by gamma (significant when relativistic).

What is your formulation of the electro-dynamic force and its strength compared to the centrifugal force?

Georges Sardin ,

I have the paper Preprint Open Challenge to the Standard Model -the Masses of Free Neu...

Using only Column potential energy, the rest mass of deuteron is found to be only -0.24 MeV/c^2 off the experimental rest mass of deuteron. That is not bad given that the charge distribution measurement is not perfect. If the experimental charge distribution of proton is not reliable, then my calculation will either be closer or farther from experiment value. The quark model is helpless in evaluating the rest mass of deuteron because it suddenly introduced so many quarks (9) and all the distances between the quarks are unknown and all the masses of the quarks are so complicated without hundreds of parameters to solve the problem. Three-body system is already very difficult to solve, no physicists have the adequate intelligence to solve a 9-body problem, forget about it.

Anyway, the two protons in deuteron are repelling symmetrically, the electron can stay in the middle to pull them together. Because the electron in the middle is closer to the two repelling protons, so the structure is stable and protons cannot get too close because they repel.

This structure is stable and no rotation is needed to maintain the structure, so the centrifugal force is an over concern. Spinning doesn't have a role in this stable structure.

Hong Du,

How would you extend your model of the static nuclear electron to the overall nuclei, starting from the electrons configuration in the tritium?

Georges Sardin ,

My model is different from Standard Model where protons and neutrons are made of quarks.

I only model neutrons as pairs of (one proton+one electron). With this model, I was able to use experimental charge distribution data to calculate the rest mass of free neutrons within 0.1 MeV accuracy, and rest mass of deuteron with about -0.24 MeV accuracy.

As of tritium, it contains two neutrons and one proton, so tritium has a mixture of three protons glued by two electrons: the two electrons will separate from each other, and the three protons will separate from each other, but the electrons as a whole will pull the three protons as a whole to form a stable nucleus. This will be a complicated model, but it will definitely be much simpler than 9 quarks interacting with each other in Standard Model.

If you don't mind, I can come up with this easy model for tritium: three protons symmetrically forms a triangular structure with two electrons sitting above and below the center of the triangle. See the attached image. If you have charge distribution data, you can give me the dimensions and I can calculate the rest mass of tritium.

Best regards.

Hong Du

You pretend having calculated the rest mass of deuteron with about -0.24 MeV accuracy.. But the electron has a high magnetic moment, so how can you get the mass of the deuteron without taking into account the magnetic interaction between the electron and the two protons? And for the tritium, the magnetic interaction between the two electrons and also with the three protons magnetic moment?

Georges Sardin ,

I don't have to pretend I know something that can be understood by high school kids. I use all experimental data, the measured charge distribution of deuteron, the measured charge distribution of proton, and then run the Coulomb potential energy. There is no artificial physics hypothesis. You can run the calculation too. If you have better experimental charge distribution of deuteron, you might get even better results than -0.24 MeV. No matter how much deviation it is, it is supported by experimental data, not something imaginary.

The high magnetic moment of electron is your hypothesis, or maybe many other physicists also like this idea. But if you support this hypothesis, you will need to get the details of it, not just something you claim to know but don't know how to calculate. At least, you should have experimental measurements of the magnetic moment of electron in deuteron. If you don't have the experimental measurement, then the hypothesis is up in the air and can be anything. I am a very strict physicist, I cannot pretend that I know something I don't know, and I cannot trust something that is not supported by experiments. BTW, how much magnetic moment of electron can contribute in your estimation? 10 eV? 100 eV? 1 keV? 100 keV? I don't think the magnetic moment can do very much. In my impression, magnetic interaction is weak unless the charge is moving highly relativistic. You will have to again find experimental evidence to prove that the deuteron is subject to high relativistic rotation. But this also needs experimental evidence.

As I presented in my paper and I attached the deuteron model in here, deuteron doesn't need to rotate, the reason is very simple: the two protons are pushed outside, they are only pulled together by the electron in the middle. The electron doesn't need to move around, it stays still between the two protons. This static model gives the lowest possible energy for the whole system. If you believe ground state with the lowest energy is more stable, extra movement of proton and electron will simply cause excited state, and the extra energy will easily be thrown out as radiation. So there is no need to rotation in this configuration, so the rotation and electron motion is completely unnecessary.

The tritium model is the same, just find out the lowest possible energy to tie the three protons together with electrons. Protons and the electrons don't need to rotate. If they do, that is extra energy that causes the tritium nucleus to radiate energy.

If you have experimental charge density data for tritium, just show it here to see if my model is reasonable, and I can use it calculate the rest mass of tritium. I like experimental data, not something that is imaginary and unknown.

Best regards.

Hong Du

You say: “I like experimental data, not something that is imaginary and unknown”. The magnetic moment of the electron and the proton are not imaginary and if you do not care to take into account their mutual interaction then your approach is partial. You also say: “In my impression, magnetic interaction is weak unless the charge is moving highly relativistic”, but I am referring to their intrinsic magnetic moment, not the magnetic moment from the electron motion. Since you argue that your model is based on the charge distribution, you should therefore show graphically the charge distributions you are using and their correspondence with the experimental ones, to evidence that they are not flawed. Anyway I am not wishing to pursue with this sterile discussion.

Georges Sardin ,

What you said is true if I do not care to take into account the magnetic moment of the electron and proton. Without the magnetic momentum of electron and proton, I have very good result, that is a fact, you can run the calculation too.

But how much is the magnetic interaction? Very weak. If you run the hydrogen atom energy calculation, you will see that the primary energy is determined by the Coulomb potential energy and kinetic energy, the spin-orbital coupling is only a secondary perturbation. For example, the 1S and 2P energy levels differ by 10.2 eV, but the spin-orbital coupling contributes only 4.5x10^-5eV. The spin-orbital coupling is caused by the intrinsic magnetic moment. You see the very weak secondary correction. If you believe magnetic moment can dominate deuteron or tritium interactions, you will have to overcome the hurdle of weakness of magnetic interaction. Don't know if you can come up a mechanism to offer the theoretical possibility.

As of the graph my model, I have attached screenshot in my prior post. Here I have attached the charge density of your paper at https://hal.science/hal-01985930/document. I didn't know that you had this paper when I wrote my paper to challenge the Standard Model, but you see that my charge distribution model is very similar to yours, and your paper's charge density is referring to the experimental data by [1] Nejc Košnik and Simon Širca, Structure of the deuteron (Department of Physics, University of Ljubljana, Nov. 17, 2004). If you compare my model and your graph, you will see that mine is very close to the experimental result. You should see the power of my model and theory, I don't even have to do expensive experiment and yet I can come up with good model and plus I can calculate the rest mass energy of deuteron within -0.24 MeV. Of course, when experiments improve, my result will certainly improve.

My theories are rarely flawed, because I am strict and I don't allow unknown imaginary components in my theories. On the contrary, I see so many theories are flawed and heading in the wrong directions and will not provide any good results except endless papers.

I will show that I always use experimental results for calculations, I never use imaginary numbers in my model. Here are the examples that I used to obtain the -0.24 MeV/c^2 rest mass difference compared to experimental value for deuteron.

The proton radius of 0.842 fm is given by Pohl R, Antognini A, Nez F, Amaro FD, Biraben F, et al. (July 2010). "The size of the proton" (PDF). Nature. 466 (7303): 213–216. Bibcode:2010Natur.466..213P. doi:10.1038/nature09250. PMID 20613837. S2CID 4424731

The deuteron radius of 2.128 fm is given by "Deuteron rms charge radius Physics.nist.gov". U.S. National Institute of Standards and Technology. Retrieved 18 September 2020.

https://en.wikipedia.org/wiki/Coulomb%27s_law#In_relativity leads to F = gamma e^2/r^2 vs gamma mvr in a circular orbit

The answer of year 2023: being a composite ("compound") of two protons and one negative pion, the deuteron should be symmetrical.

If the similar question why 2He nucleus does not exist is also interested for you, please, read the answer of year 2023: being able to bind only by negative pions, two protons cannot bind alone.

To see why it is so, you can with following links:

Poster The Proton - Pion Model of Neutron and Nucleus [discovered in 2023]

Conference Paper Application of the Compounds Identification Approach Adopted...

Presentation Application of the Compounds Identification Approach Adopted...

Hong Du

The identification of proton and electron as components of nucleus by father of nuclear physics Rutherford had only one problem. Positive particle proton was in excess, so easy to identify. The statement that negative particle is electron was opinion of Rutherford, contradicting experimental data.

The fantasy that relativistic tricks can help contradicts to the energy conservation law. The fact is following - as proton and electron is lighter then neutron, there is no mass to produce binding energy, so this pair cannot be components of neutron as composite ("compound"). Also spin 1/2 of electron contradicts to difference in spins of neutron and proton equal to 0.

The correct answer is following: negative particle must be identified correctly. The negative pion has correct spin, mass and SU(3) symmetry to be this negative particle. In that case all three kinds of charged particles participates in atom formation - baryons by lightest member proton, mesons by lightest member negative pion, electron by lightest member electron. Without strange ban for mesons to participate growing from as well strange the proton-neutron model.

Pavel V. Kudan ,

The biggest barrier to recognize the modeling of a free neutron as a proton and a electron is the positive binding energy as you pointed out (proton+electron lighter than neutron).

The barrier is caused by the defects in existing theories about the total energy. The total energy should be absolute as in the energy-momentum relation where (Etotal)^2=(mc^2+V)^2+(pc)^2. When electron is attracted by the proton, electron takes the majority of the potential energy, and when the distance is much smaller than the radius of hydrogen atom and close to 0.8 femtometers, potential energy gets very negative reaching -1.8 MeV, it surpass the rest mass of 0.511 MeV of electron, so the electron has a absolute rest mass of abs(0.51-1.8MeV)=1.29MeV. The electron in free neutron has a total energy of 1.29 MeV instead of 0.51 MeV, an increase of 0.78 MeV comared to 0.51 MeV. This means that the binding energy of the electron is positive at 0.78 MeV in free neutron. If you recall the beta decay of free neutron energy of 0.782 MeV, you will be surprised how well the modeling agrees with experimental data! I didn't use any artificial particles such as pion. It also means that some high energy particles in particle physics might just be electrons pushed too close to the proton!

You might disagree that negative energy cannot be real. But you can think of the electron with 0.78 MeV kinetic energy been accelerated and collides with proton and the electron stops somewhere around 0.8 femotmeters and stayed there to form a free neutron. Remember negative potential energy cannot be released in experimental observations if you don't trust the use of absolute energy. If you use the concept of absolute energy, then things will be much easier. When electron approaches the proton closer, the absolute energy can become bigger, but it cannot be too big, otherwise the absolute energy will exceed the available energy, so the conservation of absolute energy prevents the electron from getting too close to the proton. This means, the electron is effective repelled by the proton if too close.

Hong Du

So, proton must take somewhere a lot of enegry just for the honor to take electron as partner? Where is this source of free energy, probably, we can took it as well instead of paying for electricity?

All is easy, neutron is less energetically favorable object then sum of proton and neutron, while compound is more energetically favorable object then sum of components. The profit is binding energy.

Before 2023 it could be useful to use such concepts with electron to have at least approximatelly correct idea of composite neutron. After 2023 you can use negative pion instead to make correct calculations.

Because not electron nor neutron, but negative pion now is identificated as a partner of proton in formation of neutron and nucleus.

Pavel V. Kudan ,

The electron takes a lot of energy (0.782 MeV) to get close enough to proton to temporarily live as a free neutron. The energy is not free, either from collision or from gamma ray or any other physical process such as nuclear decay processes.

The fact that a free neutron always decays into a proton and electron says a lot as an experimental evidence that neutron can be modeled as a proton and an electron. There is no doubt about this. The only problem is that people don't know how to explain why electron has positive 0.782 MeV binding energy with proton to form a neutron. As I have explained in my prior post, the absolute energy is the key to overcome this barrier.

I don't mind if you use negative pion to model neutron. But you need to solve the following problems: the size of the model, the experiment charge distribution of the model and the rest mass of the model. If you can solve all of the problems as the proton-electron model does, then your model is a decent one. But more importantly, you will need to solve the decay measurement problem: the decayed proton can only take very little amount of kinetic energy, pion is much heavier (>130MeV/c^2) than electron (0.51 MeV/c^2), it will kick decayed proton much faster than experimentally observed proton kinetic energy. This is a major issue of your model.

Best regards,

Hong

Consider the neutron to be an isomeric state* of the femto-hydrogen atom. I call that "proton plus (>100 MeV) relativistic electron" a "fat" neutron (an atomic-mass ~1, neutral, body about 10 fermi in diameter). The proton would be a more stable isomer yet.

A (slightly) more conventional view would be to consider the proton+deep-orbit-electron to be the atomic-hydrogen ground state with both normal atomic-H and neutrons to be excited (isomeric?) atomic states.

* Occasionally the half-lives of nuclear isomers are far longer than 1e-7 seconds and can last minutes, hours, or years. For example, the 73Ta nuclear isomer survives so long (at least 1015 years) that it has never been observed to decay spontaneously.

Andrew Meulenberg

Please, study me to get over 100MeV from nothing like relativistic electron for economy on electricity.

The negative pion is particle with chare -1, mass 139.6 MeV and correct spin 0. Use it for calculation.

Please, stop torment electron, it is already have job in the atom - being on the orbit, making chemical properties. It cannot work at 2 different jobs at one time.

Pavel V. Kudan

The "nuclear stark effect"* (from the relativistic near-field Coulomb potential of a 5 fm electron orbit) spreads charged quarks further apart and therefore reduces the nucleon-mass energy. Thus, the >100 MeV kinetic energy of a deep-orbit electron comes from the nucleon mass. While not measurable, the increase in KE of an atomic electron along with the emitted photon energy also comes from the nuclear mass (potential energy) in the same manner. (Everybody recognizes the atomic mass/energy loss when the photon leaves; but, no-one seems to ask where the other 1/2 of the Coulomb-potential energy (that accelerates the decaying electron) comes from. There is no "free ride".

Even without using quantum mechanics, it is possible to find a classical, relativistic, electron deep orbit (at KE ~ 100 MeV). Thus, there are different orbit systems for the same electron. (Of course, it is possible to call such an electron a muon to deny such capability.) I have drafted proof of that, but not yet published/posted it.

* e.g., https://scholar.google.com/scholar_url?url=https://scholar.archive.org/work/blyuitkzgbgivej3p4ml6j3nwe/access/wayback/http://cipris.eu/fileadmin/CIPRIS_items/Publications/cipris_publication_36.pdf&hl=en&sa=X&ei=kQUbZa70HPq8y9YPgcCLiAg&scisig=AFWwaebhwtag644royEuGc2pStue&oi=scholarr

Andrew Meulenberg

Andrew, in addition to free energy now I have disire to see quarks.

The proton-pion model needs no tricks like quark postulations to describe SU(3) symmetry. The proton and negative pion at least real particles.

After all if electron will take 100MeV from proton, so sum of their masses will not change. If you do not trust me, try to take 100 USD from your neighbor and then try to make him believe that he still have it and you have 200.

Pavel V. Kudan

The deep-orbit electron with a proton has a binding energy of >500keV. The neutron (as a proton + electron) would be an excited state (isomer) of the H atom. It is much more likely to decay to a free proton than to the deep-orbit electron H. It is stable when paired ( and sharing the electron) with a proton to form a femto-H molecular ion (a deuteron).

Andrew Meulenberg

As professional in identification of compounds I say you - first version of Rutherford, the proton-electron model was not exact identification. It was beginning of 20th century, they did not know particles expect for proton and electron. Identification with full information available gives the negative pion as such negative particle. That is not my will, that is fact.

Sydney Ernest Grimm

The problem was wrong identification of components of nucleus by Chadwick in 1935. Now in 2023 it was solved. You may read full presentation how it was made:

Presentation The proton - pion model of neutron and nucleus [discovered in 2023]

Neutron is not elementary particle. Nor a composite between proton and electron. It is composite between one proton and one negative pion. Beta decay is decay of negative pion to electron and antineutrino. When it composite consisting of one proton and one negative pion, negative pion decays and proton gets free.

Let me present the animation of an experiment that supports the neutron shell model through the high-angle scattering of the neutron.

The law of momentum conservation prohibits the high-angle scattering of the neutron shown in the animated graph, but it actually happens. So, is the momentum conservation unfulfilled? In fact, it doesn’t. What happens is that the neutron shell is transferred to the impinging proton in 50% of the times. This deeply favors the neutron shell model by providing a straightforward explanation to the high-angle neutron scattering.

Presentation Animation of the Low and High angle n-p scattering

To start the animation, download first the file and once edited press the F5 key

Those who have a microsoft account may get it at: n-p scattering.pptx - Microsoft PowerPoint Online (live.com)

Notes: The neutron shell is traced by an integer electric charge, not by an electron and neither by a pion. The electric charge is commonly confused with the electron, however the electron is in fact a massive quantum state of the electric charge, as well as the pion is a still heavier of its allowed quantum states. The electric charge itself has no mass, but it is always dressed with a variety of mutable quantum states that do, each of them corresponding to a given elementary particle.

The neutron shell model also favors the neutron decay, which corresponds to the sole decay of its shell, which in getting free from its core proton mutates into the quantum state of the electron, the excess of energy being expelled into an anti-neutrino.

The neutron shell model also favors the cohesion of nuclei by sharing its shell with adjacent protons.

Article Low and high angle neutron-proton scattering

Georges Sardin Yes, that was old beliefs in elementary neutron combined by old speculation about neutral pion with fantastic statement that in can appear against energy conversation low.

Reality is other. There is no neutral pions appearing and desappering in the nucleus. There is no blindly postulated elementary neutron. The partner of proton is negative pion. It should not be confused with neutral pion, those are different particle. Neutron is composite of one proton and one negative pion. There is only one type of heavy particle in nucleus - proton. To which proton in nucleus at the particular moment negative pion is closer, this one we can call neutron.

This was discovered in 2023:

Presentation The proton - pion model of neutron and nucleus [discovered in 2023]

P.S. We should not confuse not only neutral pion with negative pion, but also our beliefs with experimentally proved concepts.

Pavel V. Kudan

The difference in mass between a neutron and a proton is about 1.3 MeV. And the mass of the negative pion is 139.57 MeV. How does your neutron model explain this?

Neither the electron nor the negative pion can be wrapping a proton to form a neutron because at a radius of the order of the femtometer, which corresponds to about the size of the neutron, the gyratory speed of the electron or the negative pion around the proton would be so high that their resulting relativistic mass would be so large that the corresponding centrifugal force would be extremely huge and could not be counteracted by the comparatively much smaller electromagnetic centripetal force from the core proton.

Georges Sardin You forgot about strong interaction of baryon proton with meson negative pion inside neutron, this made your considiration on sizes naive. The energy of binding of proton with negative pion is approx 140 MeV. This makes neutron as compact as proton.

Sydney Ernest Grimm

Arayik Danghyan Yes, it is correct. Real value of strong interaction between proton and negative pion in formation of neutron is approx. 140 MeV. That means that proton loose approx. 70 MeV and negative pion loose approx. 70 MeV (which consist approx. half of mass of negative pion).

Sorry that because of wrong identification of neutron as elementary particle appox. 100 years ago you all the time believed that strong interaction is a couple ordel lower.

After 2023 have choise to accept the Nature or stay in jail of beliefs.

Arayik Danghyan

I think also that it is of use to see the full presentation of the proton-negative pion from NUCLEUS-2023 international conference:

Presentation The proton - pion model of neutron and nucleus [discovered in 2023]

From the presentation you may see for the first time what is binging enegry when we are not blinded the postulates about proton and neutron interaction from the proton-neutron model postulated by Chadwick, Heisenberg and Ivanenko.

Protons does not attract each other at all. The glue is negative pions. Negative pion have a chose to bing with proton or to decay to electron and antineutrino. As protons-antiprotons pair annihilates totally and positive pion - negative pion pair annihilates totaly, as well proton and negative pion being not symmetrical to each other may releaze only the part of the energy of annihilation.

That is obvious explanation of 140 MeV energy releaze which is out of possibility of electromagnetic interaction.

After all, why are you wondered by known 140 MeV strong energy effect from interaction of proton and negative pion, but not wondered by known approx 2000 MeV strong energy from interection of proton and antiproton?

Moreover, why you are not wondered by postulates that charged baryon proton must iteract with neutral baryon neutron, as well between similar particles each introduced in the proton-neutron model based, as we know in 2023, on the wrong postulation of neutron as elementary particle in 1935?

I have pointed out that the electron at neutron radius doesn’t need to rotate around the proton because it is repelled by the proton, not like the attractive force at the hydrogen atom radius. This is the key point that existing theories don’t have a clue except my extended energy momentum relation of E^2=(mc^2+V)^2+(pc)^2. Closer distance to the proton will cost more energy for the electron in neutron, so p must remain zero, no rotation. This is opposite to hydrogen atom where electron gains kinetic energy when getting closer to the proton.

You probably won’t understand it until my theory is widely acknowledged.

The electric potential energy at neutron’s radius is only 2 MeV, it cannot afford to cancel 70 MeV. So pion idea has too much trouble and decay momentum of proton doesn’t support pion’s high rest mass. Pion idea simply doesn’t go very far.

Hong Du

The reaction of proton with negative pion is known and known its result - approx. 140 MeV releasing by 2 gamma-quants direcly or via intermedite formation of neutral pion with fast decay to 2 gamma-quants.

That is known fact.

On the other hand we have your belief that strong interaction is only 2 MeV.

That is popular, but blind belief.

I think that facts are more important that beliefs, if, of course, we are going to see what happening exactly instead of being blinded by postulates.

Pavel V. Kudan ,

Have you found gamma rays with energy greater than 2 MeV in free neutron decay? Your estimation of 70 MeV is too large compared to experiments, nothing observed at this range in nuclear processes.

The strong interaction can definitely be modeled by electrical potential energy, because shrinking the distance can do it. Don’t fight with experimental results.

Hong Du

I cannot see you between more than 200 readers of the presentation:

Presentation The proton - pion model of neutron and nucleus [discovered in 2023]

As well it is obvious, that you read nothing about reactions known for negative pion.

What is your problem? You want argue just for argue? Sorry, I have not time for that.

Pavel V. Kudan ,

you don’t have to have time. I read the paper believe or not.

I immediately see the problems that you need to solve:

(1) no decay products from pion decay observed in free neutron decay

(2) momentum of recoiled proton from free neutron decay does not support heavy pion conjecture

(3) radius of free neutron doesn’t give 70 MeV to deduct the rest mass energy of more than 100 MeV

These are all your problems, not mine. No one wants to argue with you, that’s waste of time. But every one wants the answers, those are not in your paper.

Hong Du

I am glad that finally you read a reasearch you argue.

1) The enegry conservation law must be respected. The decay of neutron to proton, muon and muonic antineutrino requers energy to brake the strong bond between proton and negative pion inside neutron. You have to give approx. 140 MeV to neutron if you like that neutron would enjoy you with products of muonic pathway of negative pion decay. Energetically favorable in this case is only electronic pathway of negative pion decay, which we see as beta decay of neutron.

Nevertheless, you can see reaction you said, in opposite direction. See so called "muonic capture" known reaction. By this reaction muon reacts with proton to give neutron (composite of proton and negative pion) and muonic neutrino. That is energetically profitable way to see the same.

2) No conjecture of negative pion is expected at beta-decay of neutron. What is expected is release of difference between energy of strong interaction between proton and negative pion and energy of decay of negative pion by more energetically favorable electronic pathway. So, I have no idea why did you calculate conjecture of negative pion.

3) When proton reacts with negative pion, it gives approx 140 MeV of energy releazed. That is energy of strong interaction, binding energy. It means that 70 MeV are from proton mass and 70 MeV is from negative pion mass. Release of enegry gives 2 equal quants. That is partial annihilation.

One who wants to see true instead of blinding youself with postulates - no matter from other researcher or self-made, can see it in the new proton-pion model based on the results of accurate identification of components of nucleus.

Pavel V. Kudan ,

Sure your research needs reading because it separates from Standard Model that I myself am working against it. And to be frank, negative pion even matches my last name because it is made of dū. But still I have to seriously doubt it.

1) You don't see muons in free neutron decay products. It is very hard to get away with the explanation of using only favored final products. Because muon has a lifetime on the order of microseconds, and muon has a rest mass energy of about 106 MeV, and when it decays, it will generate very energetic electrons, that can be tens of MeV. None of these are observed in free neutron decay. Experimental facts must be honored to set up a model, not just blinding yourself with postulates - no matter from other research or self-made.

And also, in free neutron decay, people never mentioned muon neutrino. Show me the reference if you find something different. You need to defend your theory with more theories to eliminate the flaws in your theory.

2) If favorable outcome is the only criteria, then everything is possible. I can even say free neutron has 3 protons and 2 antiprotons and 1 electron. This will make the model unspecific.

3) Specific experimental parameters must be used to describe the model, such as the charge distribution, the radius and your potential energy, otherwise theories are baseless.

Hong Du

That is exactly what I mean. If we will replace in calculation the electron for negative pion, we will have great benefits: enouph mass to obey energy conservation law, correct spin and correct SU(3) symmetry.

On the doubts:

1) muon decay of neutron requires energy from outside. It is not possible without additional energy as not favorable energetically. Favorable is decay to electron and electronic neutrio.

But the favorable reaction in opposite direction is known. See "muon capture" in the articles. In that case muon reacts with proton to give neutron and muonic neutrino. That means that muon gives negative pion and muonic neutrino - opposite reaction to decay of negative pion and muonic antineutrino. The energy profit is from binding energy of proton with negative pion at formation of neutron.

So, we should not expect what is not energetically favorable, but may see the same in other way.

2) You had to use antiprotons in your calulations because of wrong spin of electron, I think. Now with negative pion identified as partner of proton you can benefit from reducing of set of particles needed for your calculations.

3) The fact that neutron is neutral only at general having distribution of charges is known from collision. That is not question of experiment, but question of correct calculation method how to get right magnetic momentum and charge distribution from the proton - negative pion model.

There is one more complication with the pion-proton model of the neutron from the extremely short lifetime of the pions. The neutron lifetime is of the order of 15 min while the negative pion lifetime is only of 2.60 10^-8 s, i.e. its lifetime is 3.46 10^10 time shorter than that of the neutron. This forces us to have to assume that the pion speed around the proton is high enough for its lifetime to be dilated up to at least 15 min. Such a huge lifetime dilation requires a gyratory speed of the pion around the proton of 2.99781215 10^8 m/s, i.e. 0.999962 time the speed of light, to reach a relativistic lifetime dilation of 3.46 10^10 time its original lifetime. This speed of the massive negative pion is somewhat excessive to be realistic, considering the extreme associated centrifugal force.

An alternative theoretical recourse would be to assume that the dilation of the pion lifetime would be due to an alleged strong force between the wrapping negative pion and the core proton. However, the pion strong force is, according to the Yukawa strong interaction, carried out by the neutral pions, so this would require an exchange of neutral pions between the surrounding negative pion and the core proton. However, the hypothesis of the dilation of the lifetime of the negative pion ensuing from the strong force could nevertheless appear uncertain.

So, dear reader it is up to you to ponder if you adhere to the pion-proton model of the neutron.

Georges Sardin

You have demonstrated one more example of why the pion/proton model will not work for the neutron. However, if you use the same method to test the relativistic-electron-shell model, you might find a different story. The relativistic quantum-mechanical Dirac deep-orbit model provides a relativistic gamma of >200 (with r = ~5 fm). This means that the effective electron mass (>100MeV) is close to that of the pion and, according to the Yukawa strong interaction, the Coulomb attraction between electron and proton is already comparable to the pion strong force. While this deep-orbit electron has a binding energy of >500 keV (and thus is stable), for an electron to reach an unstable deeper level (a resonance inside the proton nuclear radius?), it would require considerable extra energy (for both its extra mass- and its access-energies).

If all of the extra energy were to come from an external source, then the resulting nuclear mass would be too high for a neutron (a major problem with the pion/proton model). However, if most of the extra energy were to come from the proton-mass energy itself (via the nuclear Stark effect on the proton charge(s) from the relativistic Coulomb field of the electron), then the resulting neutron mass would only be slightly greater than that of the proton.

Andrew Meulenberg Not as well, as I do not have a habbit to develop things in the imagination, postulate and then demonstrate, especially in case when they do not correspond to energy conservation law.

The specialist in identification told you the conclusion on composition of neutron. Nor Rutherford was correct identifing electron as partner of proton in nucleus formation, neither Chadwick was correct indentifing neutron as parner of proton in nucleus formation. Correct partner is negative pion.

You may replace electron for negative pion in your calculations, in that case your calculations will correspond to result of identification of components of nucleus (which is logical way), or you can use electron in spite of electron contradicts to results of identification (if you want to spend years of your life on the wrong way)

Pavel V. Kudan

It seems that your "pion plus proton = neutron" model must either violate conservation of mass/energy and/or charge. My "relativistic electron plus proton = neutron" model does neither. It might appear to violate conservation of spin; but, only total angular momentum need be conserved and the Dirac model does that.

Andrew Meulenberg Check it by arithmetics instead of "seems" and "highly likely" and this it will be a since instead of gossips and insinuations.

When you will check, you will see that pair 1 proton + 1 negative pion as components of neutron are in full correspondance with charge conservation law (as neutral charge of "compound" is a result of compensation of +1 and -1 charges of components, as well with energy conservation law (as mass of its "compound" is lower than mass of components, the difference for binding energy) as well with spin conservation law (as 1/2 spin of "compound" is a result of summing of 1/2 and 0 spins of components).

Elecron corresponds only to charge conservation law and violates both mass conservation law and spin conservation law.

It does not make your belief in proton+electron before 2023 not correct, not at all, as before 2023 you was more correct than believers of proton+neutron and quarks because at least you believed in proton+negative particle.

But after 2023 the time of beliefs passed away. Because now we have exact identificatio of that negative particle - the negative pion.

See presentation from Russian Federal Nuclear Center with detailed information how the identification was done:

Presentation The proton - pion model of neutron and nucleus [discovered in 2023]

The neutron mass is 1.293 MeV higher than the proton one, so the net energy of the negative charged particle (whatever its identity) bound to the proton must be equal to 1.293 MeV. Instead to assume that, in fusing to the proton, the electron would preserve its identity, we assume that it undergoes a structural mutation from its original quantum state of 0.511 MeV to a higher quantum state of 1.293 MeV, and so it is not an electron as such anymore. In effect, it must be taken into account that the structure of the electron, as well as that of all elementary particles, is not rigid but can mutate from one quantized state to another, as e.g. in their spontaneous decay. Let us stress also that the neutron does not need any activation energy to decay since the sum of its decay products has a lower mass.

This viewpoint presents the great advantage of providing a unified representation of all elementary particles in conceiving them as quantum states of the electric charge. Apart from the neutron structure it has several other applications that have been uploaded in researchgate. The punctual approach to the neutron structure as an electron bound to a proton maintains unsolved their own structure, as well as that of all the other elementary particles, while our approach applies to all of them, being seen as the diversity of quantum states that can acquire the electric charge.

Let us now consider the size of the proton and the neutron. The official size of the proton is of 0.84 Fermi. On its part, the neutron size can be derived from the “classical” radius of its negative charge envelop. Since this envelop has an energy of 1.293 MeV consequently its classical radius is of 1.11 Fermi, i.e. 0.27 Fermi larger than the proton. These sizes are coherent with the known size of 1.4 Fermi of the nucleons of atomic nuclei, calculated from the volume of nuclei divided by the number of nucleons, taking into account that their derived size embraces a slight separation to each other within nuclei.

Georges Sardin For that you have to postulate that different quantum states of proton exists. To postulate isospin. That if world of fantasy instead of reality. As well you can postulate that salt NaCl consist not from Na+ and Cl-, but from Na in different quantum states or make some other as well weird postulation.

Now in 2023 it was identified that neutron is a composite particle consisting of one proton and one negative pion. And deuteron is compostite particle consisting of two protons and one negative pion.

So, the answer for your original question is following: deuteron consisting of two identical protons and one negative pion is as symmetrical as hydrogen molecule deprived of one electron consisting of two identical nucleus of hydrogen and one electron.

The proton-neutron model showed to be not correct. The derived from it the quark model too.

Pavel V. Kudan

You state: “For that you have to postulate that different quantum states of proton exist”.

This is a much fortuitous asseveration of yours, as most of your many other ones in your posts. In them you just keep repeating your credo without taking account of other people remarks. For example, how would you solve the problem of the extremely short life-time of only 26 nanosecond of the negative pion versus the neutron much longer lifetime of 15 min, or your standpoint is immune to any remark.

Your have not come up with any proof but just with dogmatic asseverations on a presumed certitude of the negative pion being a component of the neutron without providing any experimental evidence for such a definitive assurance. Science does not work on just repeating a credo, it is based on proposals until some experiment favors one of them. Reiterating that all Scientifics from the past and from the present are wrong, except yourself appears somewhat pretentious. Up to now you have not provided any proof of your predications.

You also state: “Now in 2023 it was identified that neutron is a composite particle consisting of one proton and one negative pion. And deuteron is a composite particle consisting of two protons and one negative pion”.

What laboratory has evidenced your affirmation? And how would you justify that the deuteron to be stable in spite of being sustained by a negative pion with a mean lifetime of 26 nanosecond or you just do not care about this incongruence. After the pion decay what keeps sustaining the deuteron? Even though I do not coincide with the proton-electron model of the neutron, it is nevertheless much more substantiated than yours, at least it appeals to a particle, the electron, which is stable and so free of the lifetime problems.

Georges Sardin I think you are just not like when somebody touch your favorite speculations based on not motivated postulations. I understand that you like to feel yourself like a God, but constatate that it is not science, that is a way to emprove self-estim.

If you believe that decay is something magical, I will disappoint you. There no magic in decay.

Decay happens when it is energetically favorable. Velosity of decay depends on extent of the energetical profit, and pathways of decay too.

You mention velocity of decay of the negative pion in its free state. OK, to see the full picture we have also to mention that there is two ways of decay of free negative pion.

First decay pathway is muonic. It is extremely fast and results in muon and muonic antineutrino formation. It is not possible for neutron and nuclei because binging energy of negative pion with protons is greater than profit from mounic pathway of decay.

Second decay pathway is electronic. It is slow and results in electron and electronic antineutrino formation. But it is energetically more favorable and can be competitor for binding energy is some cases. Moreover, beta decay of neutron and some nuclei is electronic pathway of negative pion decay.

For the case of neutron (a composite particle consisting of one negative pion and one proton) the binding energy is lower then profit from electronic pathway of negative pion decay. For that reason negative pion decays to electron and neutrino, proton gets free. We see this three particles as a result. The process is slow, because fast muonic pathway not possible as energetically not profitable, while electronic pathway of bounded negative pion is even more slow then electronic pathway of free negative pion.

Thus, negative pion as partner of proton explains what happens when neutron or some of nuclei beta-decays. That is corrected Rutherford-Kudan model (the proton-negative pion model).

Original not corrected Rutherford model (the proton-electron) does not explain beta decay.

All reasons for correction of the proton-electron model to the proton-negative pion model are given in the presentation which, you, obviously, do not desire to see bevore your critics in spite of the fact that this identification is madden by specialist experienced in components identification from the field in which components are identified exactly from handreds of millions of possible variants - mass spectrometry.

In my country we call that way of argumentation as - I never read Pasternak myself, but I have strong opinion on his books.

OK, at least communists forbided to citizens to read Pasternak, but who forbid to you to read presentation in which a parnter of proton identified by charge, mass, spin and SU(3) symmetry and confirmed by comparison with most important properties of neutron and nuclei?

Presentation The proton - pion model of neutron and nucleus [discovered in 2023]

Pavel V. Kudan

You come up with a potpourri of fantasist deductions, such as:

* “I understand that you like to feel yourself like a God”

* “If you believe that decay is something magical”.

Did I say that the decay was magical? To defend the incoherence of your model with respect to the negative pion lifetime by just answering that the decay is not something magical is not a very good scientific argument.

* “All reasons for correction of the proton-electron model to the proton-negative pion model are given in the presentation which, you, obviously, do not desire to see…”

but I had actually read your presentation before.

* “First decay pathway is muonic. Second decay pathway is electronic…”

But this is just text book knowledge. How this can justify your pion-proton model?

* “In my country we call that way of argumentation as - I never read Pasternak myself, but I have strong opinion on his books. OK, at least communists forbided to citizens to read Pasternak, but who forbid to you to read presentation …”.

I have already said above that I had read your presentation. I do not see what the communist affair has to do with the negative pion short lifetime problems in regard to the neutron structure and as the carrier of the strong force.

If you want me to pay credit to your model, which is just a model among many others, and not a certitude as you pretend, you will have to demonstrate that the negative pion becomes stable in being bound and that it is effectually detected in the bound state, and not just come up with elusive answers.

Unless you stop selling irrevocable certitudes and present coherent argumentations I will not keep answering your posts. Please, defend your standpoint to others.

Georges Sardin

The communists had ideology which finally brake huge country, as well you have ideology.

I explained in detail for what reason decay of neutron is limited by slow decay to electron and electronic antineutrino and why ever such slow decay is not possible for stable nuclei like deuteron. That is not my guilt that you do not understand explanation based on the energy conversation law because in your ideology the energy consevation law can be violated.

In the presentation there is full information on the question how charge, mass, spin and SU(3) symmetry with beta-decay cries to us that partner of proton is negative pion. But in your ideology, fairy tailing like old Chadwick's fairy tail about elementary neutron is more important than know facts which has a name 'just text book knowledge', with full neglectance to experienced in composition identification professianals conclusions.

The proton-negative pion model is not just model based of fairy tailing, but a result of final and exact identification of the partner of the proton. That is the reason for simplisity of explanation of phenomena observing. First before to build models, we have to identify exactly components of the nuclei. That was madden.

As well it may be interesting for you to see parallel discussion here:

https://www.researchgate.net/post/Are_quarks_theoretical_fantasies/3

Sorry that there is no sweet fairy tailing there - dust hard discussion of known facts from the point of views of several models.

Georges Sardin this paper may be the best evidence against fractional quarks I have seen so far!

I have not read the comments yet, but just want to add some value after reading your paper. Please note I am fully in support and questions/comments are merely to stimulate further thinking.

There is much overlap between your (more serious) and my (naive) model. It is rewarding to see though that I am not mad after all :). I have to say that reading your shell model somewhere around 2015 gave me a huge boost in my own thinking!

I will still respond to your question, and possibly to some of the other comments, but I wanted to comment first about your figures Fig4 and Fig5,6.

The charge distribution in Fig4 is taken from r=0, meaning it is only one half of the picture. The way you represent in Fig5,6 is then not quite right. It does not detract from your argument though! I just thought I should mention.

What I think is very important to note is that the charge distribution for both proton and neutron (Fig4) thends to zero at r=0. Whereas there may have been some argument to support a 0 for the neutron before, it can no longer hold water after someone reads your paper.

What is profound though is that the proton charge in the centre is measured 0. One would have expected it to be like a cos(x) graph, having a maximum at zero. This points that the centre of the proton is not overly positive charge, meaning there is further symmetry yet to be found and definitely some negative charges contained within!

Also, in this way, the missing 'anti-matter' is in the proton!

Francois Zinserling

You said: “This points that the centre of the proton is not overly positive charge, meaning there is further symmetry yet to be found and definitely some negative charges contained within! Also, in this way, the missing 'anti-matter' is in the proton!”

Let me suggest you my proposal at this respect as a theme of reflection about the absence of antimatter.

Preprint The electron as the fossil of the anti-proton

Georges Sardin I am not sure if I can adapt to the fossil idea.

In the 1950's there were (three?) 'partons' detected in the proton as the result of high-energy scattering experiments. As a direct result of this Gell-Man and Zweig proposed their quark theories as abstract solutions.

Note I say abstract, since they never expected quarks to be real, but just a mathematical tool. This was where we took a wrong turn down the rabbit hole. Since then quarks have become 'real' in physics even though still undetected.

Point is though, we cannot ignore that there were 'partons' detected, and therefore the proton cannot be fundamental - it must be composite.

Francois Zinserling

First of all, partons just like quarks have never been detected in free state. So, they both can be considered imaginary particles.

Now, look at the uploaded diverse atomic orbitals of the hydrogen atom. Each part of the different orbitals, have not been attributed to one electron, which would make then the hydrogen diverse orbitals be traced by a varying number of electrons, but actually being traced by a single electron. (click on the image to see it full).

Now, apply this example to the proton and assume that the three supposed partons correspond to three parts of its structural orbital, somewhat similar to a cloverleaf, but in three dimensions. Each part of the charge distribution of its structural orbital, should not be naively seen as a parton. This viewpoint presents the advantage to explain why partons are not seen in free state, they cannot since not being particles but just parts of the proton structural orbital. The same consideration can be applied to the inexistent quarks as particles.

I have already treated this issue in the article about the unitary conception of elementary particles, published in 1999, where it is said. “§ I.6. Link with quarks and partons. Within the orbital context quarks as well as partons must be reinterpreted, at best (19-23).They can no longer keep their status of particles. The orbital theory may incorporate specific aspects of the parton and the quarks, but at the cost of attributing them a different nature. From the orbital perspective, partons can no longer stand to be point-like objects forming a cloud behaving as a quasi-ideal gas and leading, e.g. to the proton.”

Article Fundamentals of the Orbital Conception of Elementary Particl...

Georges Sardin

Doesn't the Oppenheimer–Phillips process prove that the deuteron is asymmetrical?

Arayik Danghyan

Let me give you the interpretation from my structural orbital approach, which is even more straightforward than the conventional n-p description. As the deuteron gets close to the positively charged target nucleus it gets polarized in the following way: the negatively charged orbital wrapping the two core protons would be attracted by the positively charged nucleus, while the two core protons of the deuteron would instead be repelled. This would lead to a brief asymmetry of the deuteron. But let me stress that this asymmetry is fleetingly induced and does not express the state of the deuteron when not exposed to external forces.

You may be interested in seeing the uploaded animation

Presentation Animation of the Low and High angle n-p scattering

Georges Sardin I am in agreement with your model and it stands to reason that the (-) shell is shared by the protons after fusion. In other words, the neutron loses identity and both protons are now in proton/neutron supposition states. Total symmetry, and it looks very much like we are used to in basic chemistry. Perfect picture in my mind.

What I still struggle to wrap my head around (shell pun intended) is why does the neutron and proton not readily fuse? If the shell exists there should be no hurdle to overcome. Yet it still needs excitation. Do you have reasoning toward this?

Francois Zinserling

Thanks for your positive critical interest in my approach and pertinent query.

You say: “What I still struggle to wrap my head around (shell pun intended) is why does the neutron and proton not readily fuse? If the shell exists there should be no hurdle to overcome. Yet it still needs excitation. Do you have reasoning toward this?”

Well, the way I see it is that not only the attraction/repulsion of the structural electric charges should be regarded, but also the fact that the structural orbitals are quantized, which means that they are somewhat “locked” into their natural state and therefore some energy input is necessary to pull them off their structural quantized state. So, they need some excitation to transit to a different structural quantum state.