PROBLEM TO BE SOLVED: To clarify what an attraction force is like when a neutron collapses and when a neutron is generated, how much the magnitude of a revolution orbit and a rotation orbit of an electron in a star far from the earth is, and how much the magnitude of the energy of an electric photon and the energy of a magnetic photon generated by the attraction force when the neutron collapses and when the neutron is generated.;SOLUTION: It is supposed that the neutron becomes hydrogen after its collapse. The revolution orbit and the rotation orbit of the electron become larger. By this, the energy of the magnetic photon is reduced. Therefore, the attraction force is reduced. A proton of the neutron comes out of the rotation of the electron. By a phenomenon opposite to that, the proton is drawn into the rotation of the electron and becomes the neutron. It is set as 2×(traveled distance)÷K=A. However, K is set as 3×108 m this time. The energy of a star can be considered as (A) times of the energy of the earth. Consequently, the revolution orbit of the electron of the star is 1/(A) of that of the electron of the earth. The energy of the star is (A) times of that of the earth. On the basis of this perception, the state of an elementary particle of the star somewhere in the universe can be understood.;COPYRIGHT: (C)2006,JPO&NCIPI
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