From the post "Opps! Lucky Me",
\(F=-\psi\)
this expression gives the force along a radial line As we know \(\psi\to 0\) as \(r\to\infty\) as such \(F\to 0\) and we are safe from an ever expanding universe.
And \(\psi\) takes the form,
\(\psi=\cfrac{D}{r^2}\)
Consider again,
\(F_s=-\cfrac{d\,E}{d\,r}=-\psi_A\)
\(F_s\) is defined wholly by a thin spherical shell of \(\psi\) passing through point \(r\) (ie. of radius \(r\)) center at the point particle. This is different from the notion that all \(\psi\) within the sphere contributes to \(F_s\). If we have,
\(\psi_A=\psi.4\pi r^2\)
And so,
\(F_s=\psi.4\pi r^2\) and
\(F=\cfrac{F_s}{4\pi r^2}=\psi\)
(We have dropped the negative sign here; \(\psi_A\) and \(\psi\) have no direction and cannot be negative. We have lost the direction information that was in the gradient of \(E\)).
Comparing the above with \(F=\cfrac{m}{4\pi \varepsilon_o r^2}\) where \(m\) establishes the field, just as charge \(q\) establishes an electric field, is a constant inside the bounding sphere of radius \(r\),
\(F_s=\psi_A=\cfrac{m}{\varepsilon_o}\)
that,
\(\psi_A=constant\)
That the surface energy density on any bounding sphere around the particle is a constant.
Furthermore, if we change the notion that all \(\psi\) within the sphere contributes to \(F_s\), but instead only \(\psi_A\), the energy density on the surface of the sphere contributes to \(F_s\) then,
\(F=\cfrac{F_s}{4\pi r^2}=\cfrac{\psi_A}{4\pi r^2}=\cfrac{m_sc^2}{4\pi r^2}\) and
\(F=\cfrac{m_s}{4\pi \varepsilon_o r^2}\)
where \(m_s\) is the equivalent mass density of the energy density on the surface of the sphere. We have
\(\varepsilon_o=\cfrac{1}{c^2}\)
This is consistent with the post "Maxwell, Planck And Particles" dated 27 Dec 14, where \(\mu_o=4\pi\times10^{-7}\) was found to be unnecessary and that we set \(\mu_o=1\), here.
And when we compare with
\(F=G\cfrac{m_s}{r^2}\)
\(G=\cfrac{c^2}{4\pi}\)
But it is true that only the surface energy density on the thin spherical shell, just next to \(r\), effects \(F\)?
