q=[3∂V∂x+∂T∂x]x=aψ.aψ
For a spread of ψ around a center,
∂ψ∂x=∂V∂x+∂T∂x=k at a particular value of x=aψ ie.
∂ψ∂t|x=aψ=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
qaψ=2∂V∂x|aψ+k
We have maximum when ∂V∂x|aψ is maximum.
If ˙x=c is a constant at light speed, then ∂T∂x is also a constant, zero, so,
∂ψ∂x=∂V∂x
which means,
qaψ=3∂ψ∂x|aψ
and q is still quantized. What is interesting is,
qaψ=3k−2∂T∂x|aψ
If T is lowered very quickly,
∂T∂x|aψ→−∞
then
qaψ→∞
If we still insist on,
∂ψ∂x=∂V∂x+∂T∂x=k
(ie. the particle not breaking up) then,
∂V∂x|aψ→∞
Fn=∂V∂x|aψ is a force orthogonal to ∂T∂x|aψ. If the latter is around a circle, then Fn is perpendicular to the circle through its center.
But where is this force driving us? Through time.
How to slow down ψ? fres=2.870Hz! 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.