The Distribution Of $\psi$ Again

This post is wrong. The derivation for force density is wrong. Please refer to following post "Opps! Lucky Me".

Consider this,

$\psi =\cfrac { E }{ Vol } =\cfrac { 3E }{ 4\pi { r }^{ 3 } }$

where $E$ is the energy at a point $r$ from a particle, ie $E(r)$.  Energy density at $r$ is thus consider to be $E$ redistributed in a sphere of radius $r$ centered at the particle.

The force density as a result of changing $\psi$ along $r$, the radial line is,

$F_{ v }=\cfrac { d\, \psi  }{ d\, r } =-3\cfrac { 3E }{ 4\pi { r }^{ 4 } } +\cfrac { 3 }{ 4\pi { r }^{ 3 } } \cfrac { d\, E }{ d\, r } =-\cfrac { 9E }{ 4\pi { r }^{ 4 } } +\cfrac { 3 }{ 4\pi { r }^{ 3 } } \cfrac { d\, E }{ d\, r }$

The total force in such a sphere is,

$F=F_v.Vol$

and so the total flux emanating through the surface area of this sphere is,

$\phi _{ total }=\cfrac { F }{ Area } =F_{ v }.Vol.\cfrac { 1 }{ Area }$

$\phi _{ total }=-\cfrac { 9E }{ 4\pi { r }^{ 4 } } .\cfrac { 4 }{ 3 } \pi { r }^{ 3 }.\cfrac { 1 }{ 4\pi { r }^{ 2 } } +\cfrac { 3 }{ 4\pi { r }^{ 3 } } \cfrac { d\, E }{ d\, r } .\cfrac { 4 }{ 3 } \pi { r }^{ 3 }.\cfrac { 1 }{ 4\pi { r }^{ 2 } }$

And so the Newtonian force, total flux per unit area is,

$F=\phi _{ A }=\phi _{ total }\cfrac { 1 }{ Area } =\left( -\cfrac { 3E }{ { 4\pi r }^{ 3 } } +\cfrac { d\, E }{ d\, r } .\cfrac { 1 }{ 4\pi { r }^{ 2 } }  \right) \cfrac { 1 }{ 4\pi { r }^{ 2 } }$

$F=\left( -\cfrac { 3E }{ { 4\pi r }^{ 4 } } +\cfrac { d\, E }{ d\, r } .\cfrac { 1 }{ 4\pi { r^{ 3 } } }  \right) \cfrac { 1 }{ 4\pi { r } }$

which is actually,

$F=\cfrac { 1 }{ 4\pi { r } } .\cfrac { d\,  }{ d\, r } \left\{ \cfrac { E }{ 4\pi { r }^{ 3 } }  \right\} =\cfrac { 1 }{ 12\pi { r } } .\cfrac { d\,  }{ d\, r } \left\{ \cfrac {3 E }{ 4\pi { r }^{ 3 } }  \right\}$

$F=\cfrac { 1 }{ 12\pi { r } } . \cfrac { d\,\psi  }{ d\, r }$

So, the Newtonian force at point $r$ is the change of energy density along the radius at $r$, per perimeter of an enclosing circle passing through $r$, centered at the particle, with a factor of $\cfrac{1}{6}$.

But we know that $F$ follows Coulomb's inverse square law, ie

$F\propto\cfrac{1}{r^2}$

so

$\cfrac { d\,\psi }{ d\, r }=\cfrac{D}{r}$

where $D$ is an arbitrary constant,

Integration both sides,

$\psi=D.ln(r)+C$

It is likely that,

$\psi=C-D.ln(r+r_o)$

for positive $\psi$, such that

$\psi_o=C-Dln(r_o)$ when $r=0$

so that $\psi$ is not infinite at $r=0$.

And at $r=r_e$,

$\psi=0=C-D.ln(r_e+r_o)$

so that $\psi$ is bounded between $r=0$ and $r=r_e$.  An illustrative plot is made below,

The gradient of $\psi$ is such that $F$ obeys the inverse square law.

The likelihood that $\psi$ does not extend to infinity but instead ends abruptly at $r=r_e$, suggests that $F$ might not obey the inverse square law when $r\lt r_e$.  For $r\gt r_e$, we know that the flux emanating from a sphere with radius $r_e$, bounding $\psi$, is a constant, as such $F$ beyond $r_e$ obeys the inverse square law.  These two points taken together, we might have a near field and far field phenomenon.

Note:  A density value at a point of distance $r$ from a center $O$, is dealt with, first by letting the value fills a sphere of radius $r$ centered at $O$ (ie. multiplied by the total volume of the sphere).  Then the total value bounded by the volume of this sphere is redistributed onto the surface area of the sphere on a infinitely thin shell (the first division by area $4\pi r^2$).  At this point the analysis has moved from a point at $r$ to values on the surface of a sphere of radius $r$, centered at $O$.  We then calculate the value per unit area on the surface (the second division by area $4\pi r^2$.