From The Very Big To The Very Small

In the post "Opps! Lucky Me" dated 25 May 2015, we have

$F=-\psi$

or

$|F|=\psi$

where the Newtonian force $F$ is the direction towards reducing force $F$.  If the particle has less $\psi$ than the surrounding then this force points towards lesser $\psi$ along the line joining the two interacting particles.  If the particle has higher $\psi$ then its surrounding then this force points towards higher $\psi$ in the surrounding along the line joining the two particles.

What happens then beyond the extend of $\psi$, beyond $x=x_a$ when the $\psi$s of the two particles do not overlap?

$\psi$ is zero beyond $x=x_a$, but $F_{\rho}$ extends to infinity at a constant value.

Does this mean that,

$|F|=\psi=0$

between the two particles?  The calculation for the Newtonian force $F$ on each particle realized through $\psi$ or $F_{\rho}$ should result in the same force.  What went wrong?

Consider this, around a spherical shell of $\Delta x$ thickness and radius $x$.  The change in $F$ as we move through the thickness of the shell is,

$\Delta F=F_{\rho}.4\pi x^2.\Delta x$

where $F_{\rho}$ is the force density at $x$.  This is the total change is over the surface area $4\pi x^2$.  For the change in $\Delta F$ along a radial line $x$,

$\Delta F_{x}=\cfrac{1}{4\pi x^2}\Delta F_{\rho}=F_{\rho}.\Delta x$

Rearranging,

$\cfrac{\Delta F_{x}}{\Delta x}=F_{\rho}$

as the we take the limit $\Delta x\to0$,

$\cfrac{d F_{x}}{d x}=F_{\rho}$

$F_{x}=\int_0^x{F_{\rho}}dx$

$F_{x}$ the force along a radial line $x$ is also the Newtonian force $F$,

$F_{x}=F=\int_0^x{F_{\rho}}dx$--- (*)

For the relation,

$|F|=\psi$ --- (**)

to be valid beyond $x=x_a$ where $\psi=0$, we have to compare the negative values of $\psi$ of both particles beyond $x=x_a$, and so admit negative $\psi$ into our model.  Both derivations are equivalent, but in (*) we deal with only positive values of $F_{\rho}$ and in (**) we have to deal with $-\psi$ values.

We obtained the expression from the post "We Have A Problem, Coulomb's Law" date 20 Nov 2014,

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

which is the force density $F_{\rho}$ associated with an energy density $\psi$, that is a circular standing wave around a center.  The equation of this circular standing wave is given by the equation,

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

and we obtain the solution,

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

This solution points to the negative direction of $x$ as the diagram above shows.  Since the square root term, $\sqrt { 2{ mc^{ 2 } } }$ in the solution allows for both positive and negative values, we admit,

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

as the solution, such that $F_{\rho}$ is in the positive $x$ direction.  (This corrects the miss-assignment previously.)

From above,

$F=\cfrac{1}{4\pi x^2}\int_0^x{F_{\rho}}dx$ 

$F=\cfrac { -i\, G\sqrt { 2{ mc^{ 2 } } }  }{ 4\pi x^{ 2 } } \int _{ 0 }^{ x }{tanh\left( \cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } (x-x_{ z }) \right)  } dx$

Since $\psi$ is positive only up to $x=x_a=2x_z$ and zero afterwards, we take limit of the integral to be from $x=0$ to $x=2x_z$,

$F=\cfrac { -i\, G\sqrt { 2{ mc^{ 2 } } }  }{ 4\pi x^{ 2 } } \int _{ 0 }^{ 2x_z }{tanh\left( \cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } (x-x_{ z }) \right)  } dx$

$F=\cfrac { -i\, 2{ mc^{ 2 } } }{ 4\pi x^{ 2 } } \left[ ln(cosh(\left( \cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } (x-x_{ z }) \right) ) \right] _{ 0 }^{ 2x_{ z } }$

which is equivalent to,

$F=\cfrac { -i\, 2{ mc^{ 2 } } }{ 4\pi x^{ 2 } } \left[ \psi(x) ) \right] _{ 0 }^{ 2x_{ z } }$

But nonetheless,

$F=-\cfrac { i\, 2{ mc^{ 2 } } }{ 4\pi x^{ 2 } } \left[ ln(cosh(\left( \cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } (x_{ z }) \right) )-ln(cosh(\left( \cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } (-x_{ z }) \right) ) \right]$

$F=-\cfrac { i\, 2{ mc^{ 2 } } }{ 4\pi x^{ 2 } } .0$

$F=0$!!!

What happened!  Obviously since $F$ comes from a 2^nd order differential equation,

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

from the post "Not Exponential, But Hyperbolic And Positive Gravity!" dated 21 Nov 2014, there can be a two constants of integration, ie

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

and the result of integration above is instead,

$F=\cfrac { -i\, 2{ mc^{ 2 } } }{ 4\pi x^{ 2 } } \left[ \psi(x) ) + Cx\right] _{ 0 }^{ 2x_{ z } }$

$F=\cfrac { -i\, 2{ mc^{ 2 } } }{ 4\pi x^{ 2 } }.2C{ x_{ z } }$

which is unit dimension consistent.  We drop $-i$,

$F=\cfrac{mCx_zc^2}{\pi x^2}=\cfrac{mCx_ac^2}{2\pi x^2}$

$F$ acts along a radial line $+x$, and $x_a=2x_z$ is the radius of the particle.

Comparing this with the expressions,

$F=\cfrac{q}{4\pi\varepsilon_o x^2}$  and $F=G\cfrac{m}{x^2}$

$\varepsilon_o=\cfrac{1}{4Cx_zc^2}=\cfrac{1}{2Cx_ac^2}$

where $q$ is considered inertia equivalent to $m$.  In this particular case $x_a$ is the radius of the charge.  This implies that $\varepsilon_o$ is different for proton and electron.

And,

$G=\cfrac{Cx_zc^2}{\pi}=\cfrac{Cx_ac^2}{2\pi}$

where $x_a$ is the radius of the planet in consideration.  $G$ is then not a universal constant, but has a factor that is the radius of the planet.

From the very small to the very big.