And $G$ Came Last

From previously, "Pinch, Pull And Let Go" dated 08 Jul 2016,

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

$KE$ is as defined in the space dimension $\cfrac{1}{2}mv^2$

$F.c=\cfrac{1}{2}m\cfrac{dv^2}{dt}$

with $v=c$

$F.c=m.c$

with $m=\cfrac{1}{c}$

$F.c=\cfrac{1}{c}.c=1$

However, force density $F$ was derived as,

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

where we make $x_z=0$.  When $m=\cfrac{1}{c}$,

$F=i\sqrt{2c}.G.tanh\left(\cfrac{G}{\sqrt{2c}}(x)\right)$

Is it possible logically that,

$F.c=\sqrt{2c}.G.tanh(\theta_{\psi}).c=1$

from which,

$G=\left(\cfrac{1}{c}\right)^{3/2}.\cfrac{1}{\sqrt{2}.tanh(\theta_{\psi})}$

where $\theta_{\psi}=\cfrac{G}{\sqrt{2c}}(a_{\psi})$

$\cfrac{1}{c}$ is the inertia of a particle in the time dimension.

$G$ is not the gravitational constant, but the constant of proportionality in the derivation for $F_{\rho}$.

Is $\cfrac{1}{c}$ the inertia of a basic particle or a big particle of $n$ constituent basic particles?

$\cfrac{1}{c}$ was derived with the consideration of $F_{\rho}$, both basic particles and big particles with $n$ constituent basic particle shares the same $F_{\rho}$ expression.  Both basic particles and big particles have the same inertia $\cfrac{1}{c}$.  When $F_{\rho}$ is scaled by $m$,

$F_{\rho}\to m.F_{\rho}$

then inertia,

$\cfrac{1}{c}\to \cfrac{m}{c}$

$m.F_{\rho}$ is due to $m$ distinct basic particles or $m$ distinct big particles.

$\psi$ is energy density without material mass.  Inertia in the time dimension is the resistance to change in energy as expressed as,

$\Delta E=\cfrac{d\,KE}{dt}$

Both basic particles and big particles although with different $a_{\psi}$ are made of the same $\psi$.  They both have the same resistance to a change in energy in the time dimension, as such they both have the same inertia.  With $m$ distinct particles, each will present the same resistance, as such there is $m\,\,\,times$ the individual inertia to a change in energy.

Good night.