From the post "Inertia In Maths",
\(\ddot { x } \cfrac { \partial \, \psi }{ \partial \, t } =\cfrac{c^3}{\sqrt{2}}(\cfrac{\partial\,F}{\partial\,x})e^{-i\pi/4}\)
where it was required that \(\dot{x}=c\), as in the case of charge.
Differentiating with respect to \(x\)
\( \ddot { x } \cfrac { \partial ^2\psi }{ \partial \, x\partial \, t } =\cfrac { c^{ 3 } }{ \sqrt { 2 } } (\cfrac { \partial ^{ 2 }F }{ \partial \, x^{ 2 } } )e^{ -i\pi /4 }\) --- (1)
The wave equation from the post "Standing Waves, Particles, Time Invariant Fields", is
\( ic\cfrac { \partial ^{ 2 }\, \psi }{ \partial \, x^{ 2 } } =\cfrac { \partial ^2\psi }{ \partial \, x\partial \, t_{ c } } \)
assuming a time invariant field, and
\(t_c=\cfrac{1}{\sqrt{2}}t.e^{-i\pi/4}\)
So,
\(ic\cfrac { \partial ^{ 2 }\, \psi }{ \partial \, x^{ 2 } } =\cfrac { \partial^2 \psi }{ \partial \, x\partial \, t_{ c } } =\sqrt { 2 } \cfrac { \partial^2 \psi }{ \partial \, x\partial \, t } e^{ i\pi /4 }\)
Substitute into (1),
\( \ddot { x } ic\cfrac { \partial ^{ 2 }\, \psi }{ \partial \, x^{ 2 } } ={ c^{ 3 } } (\cfrac { \partial ^{ 2 }F }{ \partial \, x^{ 2 } } )\)
But \(F=m\ddot{x}\),
\( \cfrac{F}{m}\cfrac { \partial ^{ 2 }\, \psi }{ \partial \, x^{ 2 } } =-i c^{ 2 } (\cfrac { \partial ^{ 2 }F }{ \partial \, x^{ 2 } } )\)
Since, \(\cfrac { \partial \, \psi}{ \partial \, x }=F\)
\( \cfrac { \partial ^{ 2 }\, \psi }{ \partial \, x^{ 2 } } =\cfrac { \partial F }{ \partial \, x } \)
\(F\cfrac { \partial \, F }{ \partial \, x } =-i mc^{ 2 } (\cfrac { \partial ^{ 2 }F }{ \partial \, x^{ 2 } } )\)
\( \cfrac { 1 }{ 2 } \cfrac { \partial \, F^{ 2 } }{ \partial \, x } =-imc^{ 2 }(\cfrac { \partial ^{ 2 }\, F }{ \partial \, x^{ 2 } } )\)
\(F^{ 2 }= -2i{ mc^{ 2 } }\cfrac { \partial F }{ \partial \, x } \)
\( \int { 1 } d\, x=- 2i{ mc^{ 2 } } \int { \cfrac { 1 }{ F^{ 2 } } } dF \)
\(x= 2i{ mc^{ 2 } } \cfrac { 1 }{ F }\)
\( F=2\cfrac { mc^{ 2 } }{ x } e^{i\pi/2} \)
The unit dimension of this expression is consistent, but it is not Coulomb's Law \(\cfrac{1}{r^2}\).
If we replace \(m\) with \(q\),
\( F=2\cfrac { qc^{ 2 } }{ x } e^{i\pi/2} \)
where the charge is at velocity \(\dot{x}=c\).
\(\psi\) was energy density defined to be energy per unit volume, as such \(F\) above is the force per unit volume. In normal flux formulation such as coulombs law, force is flux, \(Flux_A\) per unit surface area, This formulation for flux is wrong, please refer to post "Wrong Wrong Wrong" dated 25 May 15.
\( Flux_A=F.\cfrac{4}{3}\pi x^3.\cfrac{1}{4\pi x^2}=\cfrac { 2 }{ 3 } qc^{ 2 }e^{i\pi/2} \) or \(\cfrac { 2 }{ 3 } mc^{ 2 }e^{i\pi/2}\)
which is a constant. And so the flux per surface area at a distance \(x\) from the particle is,
\(F_f=\cfrac{Flux_A}{4\pi x^2}\)
\(F_f=\cfrac { 2 }{ 3 }\cfrac{ qc^{ 2 }}{4\pi x^2}e^{i\pi/2} \) or \(F_f=\cfrac { 2 }{ 3 }\cfrac{ mc^{ 2 }}{4\pi x^2}e^{i\pi/2} \)
Which is more familiar. The factor of \(e^{i\pi/2}\) is the result of the time dimension on which the particle is travelling being orthogonal to the space dimension in which we experience \(F_f\). More importantly, the expression for \(F_f\) implies that,
\(\cfrac{1}{\varepsilon_o}=\cfrac { 2 }{ 3 }c^2\) These results are wrong
\(\varepsilon_o=\cfrac { 3 }{ 2}\cfrac{1}{c^2}\)
and
\(G=\cfrac{1}{6\pi}c^2\)
All this is under the assumption that the particles manifesting gravity and the electrostatic force is travelling at light speed, \(\dot{x}=c\). Notice that \(mc^2\) in the expression for \(Flux_A\) is the kinetic energy of the particle traveling at light speed on the time axis.
