What Change $KE$?

From the post "Equating To A Indefinite Integral" dated 10 Jun 2016,

$q|_{a_\psi}=2\dot{x}F_{\rho}|_{a_\psi}$

and at a distance $r$, $r\gt a_{\psi}$ from the center of the field,

$\cfrac{1}{4\pi r^2}\cfrac{q|_{a_\psi}}{\varepsilon_o}=\dot{x}F_{\rho}|_{r}$    defined this way

where $\dot{x}=v$.

This was Coulomb's inverse square law, as a constant power is divided over the surface area of a sphere at radial distance, $r$ from the center of the field.

So, with

$\cfrac{1}{\varepsilon_o}=2c^2ln(cosh(\theta_{\psi}))$ and,

$F_{\rho}|_{r}=\cfrac{F_{\rho}|_{a_\psi}}{4\pi r^2}.\cfrac{2}{\varepsilon_o}$

here the factor $\cfrac{2}{\varepsilon_o}$ behaves like surface area,

${F_{\rho}|_{a_\psi}}.\cfrac{2}{\varepsilon_o}=\Psi_c$

is the total flux emanating from a sphere of surface area of $4\pi a_{\psi}$, at a distance $a_{\psi}$ from the center of the field.  So that, $F_{\rho}|_{r}$ is simply,

$F_{\rho}|_{r}=\cfrac{\Psi_c}{4\pi r^2}$

And when $\dot{x}=v=c$, we have,

$\cfrac{1}{4\pi r^2}\cfrac{q|_{a_\psi}}{\varepsilon_o}=F_{\rho}|_{r}.c$

This is different from Coulomb's

$\cfrac{1}{4\pi r^2}\cfrac{q}{\varepsilon_o}=F$

the force in a field per unit charge.

From Newton force,

$F=ma=m\cfrac{dv}{dt}$

An incremental work done by this force,

$F.\Delta x=m\cfrac{dv}{dt}\Delta x$

If we allow $m$ to change with velocity, ie to associate the factor $32\pi^4$ in the expression,

$32\pi^4v^2_x+v^2_t=32\pi^4c^2$

that originated from considering the different in speed limits in the time dimension and the space dimension and that energy is conserved across the two dimensions.  At the speed limit, a particle passes over to the orthogonal dimension; in space, light speed brings a particle to the time dimension at $v_t=0$.  So, we have,

$F.{ \Delta x }=\left\{ m(v_{ p })+\Delta m(v_{ p }) \right\} \cfrac { dv_{ p } }{ dt } { \Delta x_{ p } }$

where $m=m(v_p)$, the particle mass is a function of its velocity, and the particle is displaced by $\Delta x_p$ instead.  $\Delta m(v_p)$ is due to a change $v_p$, $\Delta v_p$ with $\Delta x_p$,

$F.{ \Delta x }=m(v_{ p })\cfrac { dv_{ p } }{ dt } { \Delta x_{ p } }+\Delta m(v_{ p })\cfrac { dv_{ p } }{ dt } { \Delta x_{ p } }$

$F.{ \Delta x }=m(v_{ p })\cfrac { dv_{ p } }{ dt } { \Delta x_{ p } }+\cfrac { dm(v_{ p }) }{ dv_{ p } } \Delta v_{ p }.\cfrac { dv_{ p } }{ dt } { \Delta x_{ p } }$

$F.{ \Delta x }=m(v_{ p })\cfrac { dv_{ p } }{ dt } { \Delta x_{ p } }+\cfrac { dm(v_{ p }) }{ dt } \Delta v_{ p }.{ \Delta x_{ p } }$

Over time $\Delta t$,

$F.{ \cfrac { \Delta x }{ \Delta t }  }=m(v_{ p })\cfrac { dv_{ p } }{ dt } { \cfrac { \Delta x_{ p } }{ \Delta t }  }+\cfrac { dm(v_{ p }) }{ dt } \cfrac { \Delta v_{ p } }{ \Delta t } .{ \Delta x_{ p } }$

as $\Delta t\rightarrow 0$

$\cfrac{\Delta x_p}{\Delta t}=v_p$,  $\cfrac{\Delta v_p}{\Delta t}=a$

$F.{ \cfrac { \Delta x }{ \Delta t }  }=m(v_{ p })v_p\cfrac { dv_{ p } }{ dt } +\cfrac { dm(v_{ p }) }{ dt } a.{ \Delta x_{ p } }$

since $\Delta t$ is small and $\Delta x$ is small, $a$ is a constant, basically,

$v=u+at$

$\Delta x=ut+\cfrac { 1 }{ 2 } at^{ 2 }$

$\Delta x=u\cfrac { (v-u) }{ a } +\cfrac { 1 }{ 2 } \cfrac { (v-u)^{ 2 } }{ a }$

$2a\Delta x=2(uv-u^{ 2 })+(v-u)^{ 2 }=v^{ 2 }-u^{ 2 }$

when $u=0$, the particle is subjected to the field from rest,

$2a\Delta x=v^{ 2 }$

$a\Delta x=\cfrac { 1 }{ 2 } v^{ 2 }$

So,

$F.{ v }=m(v_{ p }).v_{ p }\cfrac { dv_{ p } }{ dt } +\cfrac { 1 }{ 2 } v^{ 2 }_{ p }\cfrac { dm(v_{ p }) }{ dt }$

$F.{ v }=m(v_{ p })\cfrac{1}{2}\cfrac { dv_{ p }^2 }{ dt } +\cfrac { 1 }{ 2 } v^{ 2 }_{ p }\cfrac { dm(v_{ p }) }{ dt }=\cfrac { d }{ dt }\left\{\cfrac{1}{2}m(v_p).v_P^2\right\}=\cfrac { dKE }{ dt }$

$v$ is independent of $v_p$, it is a parameter associated with the field only.

From previously,

$c=n\sqrt { \cfrac {(32\pi ^{ 4 }-1) }{ 128\pi\theta_{\psi}ln(cosh(\theta_{\psi}))tanh(\theta_{\psi}) }  }$

$n=\left(\cfrac{\theta_{\psi}}{0.7369}\right)^3$

$c$ is the flow of energy from  the center of the field, irrespective of the velocity of a test charge in the field.  $c$ is a constant between all interacting particles with the same $n$ number of constituent basic particle.

When $v=c$, the idea of $F$ changing $KE$ directly and $F.c$ changing $KE$ directly differ by a constant.

$F.c=\cfrac { dKE }{ dt }$

$F.c$ is decreasing with distance $r$ from the center of the field because the fixed total amount of energy from the field passes through a bigger surface area $4\pi r^2$ at $r$.
...

This is still prelude to an expression for mass $m$, but correct for $F.c$ that change $KE$ directly.

$\cfrac{1}{4\pi r^2}\cfrac{q|_{a_\psi}}{\varepsilon_o}=F_{\rho}|_{r}.c\ne F_{newton}$

Two things have changed, a field is quantifies by its power output $F.c$ and the force in a field is different from the Newtonian force.