\({q}=\left[3 \cfrac { \partial V\, }{ \partial \, x } +\cfrac { \partial \, T }{ \partial \, x } \right]_{x=a_{\psi}}.a_{\psi}\)
For a spread of \(\psi\) around a center,
\(\cfrac{\partial\psi}{\partial x}=\cfrac{\partial V}{\partial x}+\cfrac{\partial T}{\partial x}=k\) at a particular value of \(x=a_{\psi}\) ie.
\(\cfrac{\partial\psi}{\partial t}|_{x=a_{\psi}}=0\)
\(k\) is a constant for stationary (in time) fields, where a particle passing through a particular point \(x\), will always have the same \(V\) and \(T\) values. We have
\(\cfrac{q}{a_{\psi}}=2\cfrac { \partial \, V }{ \partial \, x }|_{a_{\psi}}+k\)
We have maximum when \(\cfrac { \partial \, V }{ \partial \, x }|_{a_{\psi}}\) is maximum.
If \(\dot{x}=c\) is a constant at light speed, then \(\cfrac { \partial \, T }{ \partial \, x }\) is also a constant, zero, so,
\(\cfrac{\partial\psi}{\partial x}=\cfrac{\partial V}{\partial x}\)
which means,
\(\cfrac{q}{a_{\psi}}=3\cfrac { \partial \, \psi }{ \partial \, x }|_{a_{\psi}}\)
and \(q\) is still quantized. What is interesting is,
\(\cfrac{q}{a_{\psi}}=3k-2\cfrac { \partial \, T }{ \partial \, x }|_{a_{\psi}}\)
If \(T\) is lowered very quickly,
\(\cfrac { \partial \, T }{ \partial \, x }|_{a_{\psi}}\rightarrow -\infty\)
then
\(\cfrac{q}{a_{\psi}}\rightarrow \infty\)
If we still insist on,
\(\cfrac{\partial\psi}{\partial x}=\cfrac{\partial V}{\partial x}+\cfrac{\partial T}{\partial x}=k\)
(ie. the particle not breaking up) then,
\(\cfrac{\partial V}{\partial x}|_{a_{\psi}}\rightarrow \infty\)
\(F_{n}=\cfrac{\partial V}{\partial x}|_{a_{\psi}}\) is a force orthogonal to \(\cfrac { \partial \, T }{ \partial \, x }|_{a_{\psi}}\). If the latter is around a circle, then \(F_{n}\) is perpendicular to the circle through its center.
But where is this force driving us? Through time.
How to slow down \(\psi\)? \(f_{res}=2.870\,\,Hz\)! From the post "A Shield" dated 27 may 2016. On Earth, because Earth is also one big charge particle.
It is still dream land, only deeper...sleep on this,
Good night.
