$\psi$ Reaches Further In Theory

The introduction of a constant $C$, in the force density equation changes everything,

$F=-i\sqrt { 2{ mc^{ 2 } } }\,G.tanh\left( \cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } (x-x_z) \right)+C$

The associated $\psi$ after integration,

$\psi= { -i\, 2{ mc^{ 2 } } }ln(cosh(\left( \cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } (x-x_{ z }) \right) ) +Cx+A$

A sample plot of $\psi$ with varies values of $C$ is,

From the diagram, for $\psi$ to be bounded such that its integral over all positive value of $\psi$ is finite,

$0\le C\lt1$

We see that when $\psi(x=x_a)=0$, $x_a\ne 2x_z$ except when $C=0$.  In fact, $x_a\gt 2x_z$.  The far zero of $\psi$ when $C\gt0$ is further than when $C=0$.

For all valid values of $C$ where $\psi$ intersects the $x$ axis again, we have Coulombs' Inverse Square Law.

$F\propto\cfrac{1}{x^2}$

If this is the case, we do not have to consider $\psi\lt0$ at all.  $F$ is due to the interaction of $\psi$s, in both derivations of $F$.

$|F|=\psi$   and

$F=\cfrac{1}{4\pi x^2}\int_0^x{F_{\rho}}dx$

The Newtonian force $F$, from the post "From The Very Big To The Very Small",

$F=\cfrac { -i\, 2{ mc^{ 2 } } }{ 4\pi x^{ 2 } } \left[ ln(cosh(\left( \cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } (x-x_{ z }) \right) )+Cx \right] _{ 0 }^{ x_{ a } }$

$F=\cfrac { -i\, 2{ mc^{ 2 } } }{ 4\pi x^{ 2 } }\left[ ln(cosh(\left( \cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } (x_a-x_{ z }) \right) )\\-ln(cosh(\left( \cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } (x_{ z }) \right) )+Cx_a\right]$

If we define,

$N(x_a)=ln(cosh(\left( \cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } (x_a-x_{ z }) \right)-ln(cosh(\left( \cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } (x_{ z }) \right) ) )$

And we have,

$F=\cfrac { -i\, 2{ mc^{ 2 } } }{ 4\pi x^{ 2 } }\left[ N(x_a)+Cx_a\right]$ 

the expression for $F$ splits into two fields.  One field due the curve part of $\psi$, $N(x_a)$ and another due to the linear part of $\psi$, $Cx_a$.  A near and a far field, each with its own aggregated constants.  Which is which?  Others might define far field as $x\gt x_a$ and near field as $x\lt x_a$ using the boundary where $\psi(x_a)=0$ as divide.  In which case, $N(x_a)$ is then a new near field, a nearer field.

Have a nice day.