\(F=-\psi\)
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 2nd 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.
