- The first basic notion is nuclear string identifiable as a monopole flux tube. The nucleus would consist of one or more nuclear strings and they would define Hamiltonian cycle going through all vertices of the Platonic solid assignable to j-shell with angular momentum j=l +/- 1/2 and number of states N_{-}=2l or N_{+}=2l+2.
- The fact that flux tubes involve also Coulomb flux does not allow closed Hamiltonian cycles: these are possible only for electromagnetically neutral systems.
One can however eliminate one edge from the cycle. This kind of quasicycle for which the flux tube arrives point A from another Platonic solid and flows through all the remaining points of the cycle and returns to a neighboring point B and continues to a neighboring Platonic solid. This kind of quasicycles can be generated when the flux tube portions defining edges of closed cycles form reconnections: in this case however both the incoming and outgoing flux tube would correspond to missing edges. This would also allow degenerate quasi cycles with 2 vertices required by the model.
One can ask whether one should one allow Hamiltonian paths in which incoming and outgoing fluxes are not associated with neighboring vertices. Their number is obviously larger than the number of cycles.
- Both the harmonic oscillator potential used in the simplest nuclear model and Coulombic potential used in the model of atoms are characterized by a principal quantum number n such that l=1,2,..,n orbital angular momenta are realized for it.
This motivates the idea of nucleus-atom holography meaning that the protonic states of the nucleus correspond to the states of electrons of the atom. This also leads to a speculative question whether neutrinos could play the role of neutrons in atoms.
- For the first option the particles could be assigned to V vertices of the platonic solid or to the V edges of the Hamilton cycle: if the quasi-Hamilton cycles are allowed, then only the vertices are allowed. The numbers of vertices are given by V= 4,6,8,12,20 for T,O,C,I,D.
It is interesting to look in detail at the assignments of states at different n-shells.
- n=0: l=0. There are 2 states. Degenerate Platonic solid as diametrically opposite points of the sphere.
- n=1: l=0,1. There are 2+2+4 states. 2 degenerate Platonic solids + T
- n=2: l=0,1,2: 4+6 additional states to n=1-shell. Additional T and O.
- n=3: 6+8 additional states to n=2-shell. Additional O and C.
- n=6: 8+10 additional states to n=3 shell. 6 corresponds to C. 10 has no counterpart as a single platonic solid. 10 → T +O.
- n=6: 10 +12 states: 12 corresponds to I. 10→ T+O.
- For the second option one assigns particles at the centers of complementary edges which by definition do not belong to the Hamiltonian cycle. There are F-2 complementary edges.
One has F-2\in {2,6,4,18,10} for T,O,C,D,O.
- n=0: The 2 states correspond to T.
- n=1: The 2+4=6 additional states correspond to T+C:
- n=2: l=0,1,2: The 4+6=10 additional states correspond to C+O.
- n=3: There are 6+8 additional states. 6 corresponds to O but 8 corresponding to j=3+1/2 is missing. T+O would give 2+6=8 and C+C would give 4+4=8. j=3+1/2 cannot therefore correspond to a single platonic solid.
Energy shell can be defined in terms of an energy gap to the next state with a higher energy and this suggests that the discrepancy relates to the fact that l=0 state of n=3 shell are near to the energy of the highest state of the n=2.
Spin-orbit interaction comes first into mind since it distinguishes between the energies for a given value of n and comes first to mind. L•S term is vanishing and its spin-orbit interaction is therefore expected to be smallest for l=0 state. In the case of atom, the interaction energy is nonvanishing since it involves expectation value of 1/2dV/dr, where V is in the atomic case Coulomb potential, in l=0 state and gives a term proportional to 1/l which at the limit l→ 0 gives a non-vanishing net result. In the case of a nucleus, the harmonic oscillator potential would give vanishing interaction energy.
The F-2 option does not require the somewhat questionable degenerate Platonic solid but the V option works also for n=3.
- One can ask whether the notion of n-shell could allow a description in terms of Platonic solids? In atoms l=0 and l=1 shells for n=1 shells give 2+2+4 =8 states, which could be assigned to the 8 vertices of the cube. F-2=8 is not satisfied by any Platonic solid.
l=0,1,2 shells for the n=2-shell correspond to 2+ 2+4+4+6= 8+10=18 states assignable to the n=2 shell. These 18 states cannot be assigned to the vertices of a single Platonic solid. These states can be however assigned with the complementary edges of the icosahedron with F-2 =18. It would look however strange to assign the n=2 shell to complementary edges of the icosahedron and n=1 shell to the vertices of the cube.
- ^{28}O nucleus has 8 protons and 20 neutrons and is doubly magic and should be therefore stable. It has 12 surplus halo neutrons and decays to a state with 8 surplus neutrons plus 4 neutrons with a life-time about 10^{-21} seconds. The 12 surplus neutrons in the halo cannot correspond to a full shell. This could explain the short life-time.
- ^{28}O decays by emitting 4 neurons to ^{24}O with 8 surplus neutrons. This state should be rather stable.
- The reason why one cannot apply the magic nucleus rule could be that halo neutrons are different from the core neurons and must be treated separately. A possible reason is that the halo neutrons correspond to a non-standard value of h_{eff}=nh_{0}>h. This can occur also for the valence electrons of rare earth metals.
- The 12 surplus neutrons in the halo do not correspond to a full n-shell. Both V and F-2 options are doomed to fail if the stability corresponds to a full n-shell.
The ordinary 8 neutrons of ^{16}O could correspond to a full n=1 shell. 8+4=12 halo neutrons would naturally correspond to a partially filled n=2 shell having 8+4+6=18 neutrons. This does not depend on whether one has V or F-2 option.
- The 8 halo neutrons have the same quantum numbers as the full n=1 shell, which suggests stability. This conforms with the experimental findings. 4 neutrons, which correspond to j=3/2-plet could be assigned with the complementary edges of the cube but cannot form a full shell since the 6 j= 5/2-plet is missing.
For a summary of earlier postings see Latest progress in TGD.
For the lists of articles (most of them published in journals founded by Huping Hu) and books about TGD see this.
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