swansont Posted December 20, 2017 Posted December 20, 2017 14 minutes ago, Butch said: Note that in my model the neutron is actually an electron inside the charge radius of the nucleus, hence if the neutron decayed it would simply mean that a proton electron pair (or pairs) would be ejected from the nucleus. The pair would be tightly bound however the electron would be high energy, hence they would immediately dissociate, observed as the decay of a free neutron. But a proton has spin 1/2, the neutron has spin 1/2, and the electron has spin 1/2. The neutron cannot be simply a proton and an electron bound together. That model is wrong. Besides which, there is no model that has the electron in such a bound state, but discussing that is irrelevant until you address the violation of conservation of angular momentum your model has.
Butch Posted December 20, 2017 Author Posted December 20, 2017 1 minute ago, swansont said: Besides which, there is no model that has the electron in such a bound state, but discussing that is irrelevant until you address the violation of conservation of angular momentum your model has. The sum of the vectors of all of the sister atoms would agree with qm. 2 minutes ago, swansont said: But a proton has spin 1/2, the neutron has spin 1/2, and the electron has spin 1/2. The neutron cannot be simply a proton and an electron bound together. That model is wrong. They are not bound together in the atom, they are strongly interacting, each sister atom takes it place eventually as the valent electron, only when the atom decays do the electron and proton seem to be bound as they are outside the nucleus charge radius, at which time they quickly dissociate.
swansont Posted December 20, 2017 Posted December 20, 2017 5 minutes ago, Butch said: The sum of the vectors of all of the sister atoms would agree with qm. 1/2 + 1/2 cannot leave you with 1/2. Neutrons by themselves have to comply with all of the conservation laws. Hiding it in a larger system will not give you cover. In any event, an unsupported claim like this is not worth the electrons used to send it out into the internet.
Butch Posted December 20, 2017 Author Posted December 20, 2017 (edited) 32 minutes ago, swansont said: 1/2 + 1/2 cannot leave you with 1/2. Neutrons by themselves have to comply with all of the conservation laws. Hiding it in a larger system will not give you cover. In any event, an unsupported claim like this is not worth the electrons used to send it out into the internet. More simply put, if you add the vector of the sister electrons the average energy would always equal that of the energy of a single qm electron at an energy of it's greatest probability... That electron however would have a probability curve that would never allow it to be in the nucleus charge radius of the atom. This is most easily understood in the deuterium atom. 2 sister atoms one at its highest probability and it's lowest energy state, the other in the center of the nucleus charge radius, at its highest energy state with a vector 180 degrees opposite the valent sister and with a much greater magnitude such that the addition of the vectors would produce an average energy equal to that of a single qm electron at its highest probability. How is an electron created? How is a proton created? How is a neutron created? Since the valent electron has energy the addition of the vectors of the sister electrons would produce a resultant that would always place the singular electron outside of the nucleus charge radius. That is to say the combined vectors would always produce a resultant having less energy than the sister electron inside the nucleus charge radius. Edited December 20, 2017 by Butch
swansont Posted December 20, 2017 Posted December 20, 2017 49 minutes ago, Butch said: More simply put, if you add the vector of the sister electrons the average energy would always equal that of the energy of a single qm electron at an energy of it's greatest probability... That electron however would have a probability curve that would never allow it to be in the nucleus charge radius of the atom. This is most easily understood in the deuterium atom. 2 sister atoms one at its highest probability and it's lowest energy state, the other in the center of the nucleus charge radius, at its highest energy state with a vector 180 degrees opposite the valent sister and with a much greater magnitude such that the addition of the vectors would produce an average energy equal to that of a single qm electron at its highest probability. How is an electron created? How is a proton created? How is a neutron created? Since the valent electron has energy the addition of the vectors of the sister electrons would produce a resultant that would always place the singular electron outside of the nucleus charge radius. That is to say the combined vectors would always produce a resultant having less energy than the sister electron inside the nucleus charge radius. You have not solved the problems before you. Neutrons have spin 1/2, and cannot be the combination of an electron and a proton. Free neutrons do not have an orbiting electron. Hydrogen angular momentum states are not consistent with an orbit. You have no way to predict the energy states of hydrogen. In fact the implication here, with e.g. two electrons, is that each would have half of the energy. This is wildly inconsistent with the observed spectrum of deuterium. But I shouldn't have to guess at this — it should be something you already should be able to predict mathematically. Y O U R M O D E L I S W R O N G. It is wrong enough that it is probably not salvageable. If you patch one wrong behavior, you will break something else. This is not the venue to "work through" problems with the model. In speculations we expect such fundamental issues to already be addressed. You are still free to ask questions to learn about mainstream physics, but until you have a model to predict behavior of the atom, this has to stop.
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