Prelude To "What Is Inerita?"

From the previous post "Entanglement Made It So" dated 29 Jun 2016,

$v^2_x+v^2_t=c^2$

$v_x$ is in the space dimension, $v_t$ is in the time dimension, both are of different reference frame.  In the time dimension the unit of measure for velocity, displacement per unit time, changes with the unit of time that lengthens with increasing velocity in time.  We can, however, compare the maximum points in each of the dimension and expresses both in the space dimension.  From the post "Just When You Think c Is The Last Constant" dated 26 Jun 2016, for example, the maximum kinetic energy of a particle in the time dimension, with reference to the space dimension was,

$E_t=32\pi^4.\cfrac { 1 }{ 2 } mc^{ 2 }$ --- (**)

compared to

$E_{x}=\cfrac { 1 }{ 2 } mc^{ 2 }$

in the space dimension.

So, (*) above with reference to the space dimension only, could be,

$32\pi^4v^2_x+v^2_t=32\pi^4c^2$ --- (*)

that would satisfy the two extremema, $v_x=0\,\,\,v^2_t=32\pi^4c^2$ and $v_t=0\,\,\,v^2_x=c^2$

But in between these extremema, the transition of $v_x$ to $v_t$ is not in linear proportion.

If we associate the factor $32\pi^4$ with the mass of the particle, then as the velocity $v_x$ of a particle of mass $m$, increases towards light speed $c$, its mass increases to $32\pi^4m$ and its $KE$ is given by (**) with the understanding that the particle is now in the time dimension.

The point of this is prelude to find the particle's mass, $m$ from $\psi$ and light speed $c$.

For the sake of...

$\cfrac{v^2_t}{c^2}=\gamma^2=32\pi^4\left(1-\cfrac{v_x^2}{c^2}\right)$

$\gamma=4\sqrt{2}.\pi^2\sqrt{\left(1-\cfrac{v_x^2}{c^2}\right)}$

Einstein, or if you travel through time.

$F=m_ia=m_i\cfrac{dv}{dt}$

An incremental work done by this force,

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

over time $\Delta t$,

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

as $\Delta t\rightarrow 0$

$F.v=m_iv\cfrac{dv}{dt}=\cfrac { 1 }{ 2 } m_{ i }\cfrac { dv^{ 2 } }{ dt } =\cfrac { dKE }{ dt }$

where $F$ changes along the distance between the particle and the center of the field.

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

is consistent, where the input power (rate of change in energy) results in a change in kinetic energy.  From,

$F.v=\cfrac { 1 }{ 2 } m_{ i }\cfrac { dv^{ 2 } }{ dt }$

$m_{ i }=\cfrac { 2F.c }{ \cfrac { dv^{ 2 } }{ dt }  }$

From (*),

$32\pi ^{ 4 }\cfrac { dv^{ 2 }_{ x } }{ dt } +\cfrac { dv^{ 2 }_{ t } }{ dt } =0$

since $c^2$ is a constant.

$\cfrac { dv^{ 2 }_{ x } }{ dt }=-\cfrac{c^2}{32\pi^4}\cfrac { d\gamma^{ 2 } }{ dt }= \cfrac { dv^{ 2 } }{ dt }$

and we are not getting anywhere.

$m_i=-\cfrac{64\pi^4}{c}F.\left(\cfrac { d\gamma^{ 2 } }{ dt }\right)^{-1}$

This particle is not at rest in the field so, $\cfrac { dv^{ 2 } }{ dt }\ne0$ and $\cfrac { d\gamma^{ 2 } }{ dt }\ne0$.  As the particle speed increases in the field $F$, time speed decreases and so $\left(\cfrac { d\gamma^{ 2 } }{ dt }\right)\lt 0$.

At light speed,

$\lim\limits_{v\to c}{F.\left(\cfrac { d\gamma^{ 2 } }{ dt }\right)^{-1}}\rightarrow-1$

as both $F$ and $\cfrac{d\gamma^2}{dt}$ is $\cfrac{dv^2}{dt}$ dependent.

$\lim\limits_{v\to c}m_i=\cfrac{64\pi^4}{c}$

This is the mass of the particle at light speed, $c$.

And boldly we are at where no rational man has gone before, because

$\lim\limits_{v\to c}{F.\left(\cfrac { d\gamma^{ 2 } }{ dt }\right)^{-1}}\rightarrow-\cfrac{c}{64\pi^2}m_i$

and

$\lim\limits_{v\to c}m_i=m_i$

It seems that rest mass is the mass at light speed.  The speed limits in the space dimension and the time dimension are different; $c$ and $4\pi^2\sqrt{2}$ respectively.

Still, how to obtain $m_i$ from

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

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

???

Once $c$ is defined and given a unit, $ms^{-1}$, $kg$ is also defined.