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 \(4\pi r^2\).
