\(ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } } }a_{\psi\,c}))=\cfrac{1}{4}\)
then it takes four basic particles to balance the charge of one big particle with,
\(ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } } }a_{\psi\,\pi}))=1\)
This might explain the stable four electron configuration, but a charge with one-fourth the charge of an electron is no longer an electron.
In the case of paired orbits, the orbiting particles are at a phase angle of \(\theta_p=\cfrac{\pi}{4}\) when they are on the same plane, two at the intersections of the orbits and when their orbits bring two of them on a perpendicular plane. In the last case, a third circle can be drawn perpendicular to the two orbits, that intersects all the revolving particles. All particles are at the same speed.
In the case of four orbitals around a single nucleus, it is possible that all particles passes through one of the two intersections of their orbits (taken in pairs) and are \(\theta_p=\cfrac{\pi}{4}\) with each other. At any time during their revolutions, a circle can be drawn through all particles and they rest at \(\theta_p=\cfrac{\pi}{4}\) phase to each other.
Maybe. The revolving particles might not even be in fixed circular orbits but are always part of a circular orbit.
So, a stable four "electron" configuration and a stable eight "electron" configuration are for different reasons (c/f "Stable Eight" dated 08 Jul 2016).
And it rained in July...
