Not Exponential, But Hyperbolic And Positive Gravity!

From the post "We Have A Problem, Coulomb's Law", we formulated the expression for $F$,

$\cfrac{F}{m}\cfrac { \partial ^{ 2 }\, \psi  }{ \partial \, x^{ 2 } } =-i c^{ 2 } (\cfrac { \partial ^{ 2 }F }{ \partial \, x^{ 2 } } )$

but  $\cfrac { \partial ^{ 2 }\, \psi  }{ \partial \, x^{ 2 } } =\cfrac { \partial F }{ \partial \, x }$

So,

$F\cfrac { \partial \, F }{ \partial \, x } =-i mc^{ 2 } (\cfrac { \partial ^{ 2 }F }{ \partial \, x^{ 2 } } )$

$\cfrac { 1 }{ 2 } \cfrac { \partial \, F^{ 2 } }{ \partial \, x } =-imc^{ 2 }(\cfrac { \partial ^{ 2 }\, F }{ \partial \, x^{ 2 } } )$ --- (*)

If we consider a constant term after the first integration,

$F^{ 2 }-A=-i2{ mc^{ 2 } }\cfrac { \partial F }{ \partial \, x }$

This term corresponds to loss in $F$ as it progresses along $x$.  This is equivalent to $F$ being in a medium with mass, ie. space has inertia.

$\int { 1 } d\, x=-i2{ mc^{ 2 } }\int { \cfrac { 1 }{ F^{ 2 }-A }  } dF$

$x=i2{ mc^{ 2 } }\cfrac { 1 }{ \sqrt { A }  } archtanh(\cfrac { F }{ \sqrt { A }  } )+x_z$

$-i\cfrac { \sqrt { A } (x-x_z) }{ 2{ mc^{ 2 } } } =archtanh(\cfrac { F }{ \sqrt { A }  } )$

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

From (*) it is possible that

$F^{ 2 }=-2i{ mc^{ 2 } }\left(\cfrac { \partial F }{ \partial \, x }+D^2\right)$

where $D^2$ is the constant of integration,

$F^{ 2 }+i2{ mc^{ 2 }D^2 }=-i2{ mc^{ 2 } }\cfrac { \partial F }{ \partial \, x }$

and so,

$A=-i2{ mc^{ 2 } }D^2=e^{i3\pi/2}.2{ mc^{ 2 } }D^2$,

$\sqrt{A}=e^{i3\pi/4}D\sqrt{2{ mc^{ 2 }}}$,

$-i\sqrt{A}=e^{-i\pi/2}.e^{i3\pi/4}D\sqrt{2{ mc^{ 2 }}}=e^{i\pi/4}D\sqrt{2{ mc^{ 2 }}}$

Therefore,

$F=e^{ i3\pi /4 }D\sqrt { 2{ mc^{ 2 } } } .tanh\left( \cfrac { { D } }{ \sqrt { 2{ mc^{ 2 } } }  }( x-x_o).e^{ i\pi /4 } \right)$

If $G=D.e^{ i\pi /4 }$ is real then

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

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

where $e^{i\pi/2}=i$ is interpreted as a phase lag in time. An illustrative plot of $-F$, which is often the case where minus gravity, $-g$ is shown,

This solution suggests that there is a zero gravity value in the space around a body at some point $x_z$, beyond this point gravity is positive and pushes other mass away.

The series expansion of $\tanh(x)$ is given by,

$tanh(x)=x-\frac { 1 }{ 2 } x^{ 3 }+\frac { 2 }{ 15 } x^{ 5 }-\frac { 17 }{ 315 } x^{ 7 }...\left| x \right|$

$tanh(x)\approx x$.  This is a different result from when $A=0$.

This force, $F$ was derived from energy density, as such it is a force per unit volume, ie Nm^-3.  To obtain the normal Newtonian formulation, $F_g$,

$F_g(x)=F\cfrac{4}{3}\pi x^3\cfrac{1}{4\pi x^2}\cfrac{1}{4\pi x^2}=F\cfrac{1}{12\pi x}$

Note: $arctan\left( x \right) =i\frac { 1 }{ 2 } ln\left( \cfrac { 1-ix }{ 1+ix }  \right)$