And we have a problem,
\(F=\cfrac{1}{12\pi r}\cfrac{d\,\psi}{d\,r}\)
\(RHS=m^{-1}.Jm^{-3}m^{-1}\)
\(J=kgm^2s^{-2}\)
\(RHS=kgm^{-3}s^{-2}\)
\(LHS=kgms^{-2}\)
\(m^{-4}\) missing?
WTF! This also means that the post "Flux It" dated 21 Nov 14 is also wrong.
But \(\varepsilon_o\) to the rescue! It is possible to introduce a constant \(\varepsilon_o\) to remove the inconsistency in unit dimensions and at the same time introduce the unit for coulomb charge \(C\). But it is first, more important to realize that \(\psi\) due to a point source is distributed uniformly in a sphere with the point source as center. The force due a change in \(\psi\) along a radial line, at \(r\) distance from the center, is also divided evenly over the surface area of the sphere \(4\pi r^2\). As such the force at \(r\), along a radial line, is the force due to \(\psi\) per unit surface area of the sphere of radius \(r\). In this way, strictly speaking,
\(LHS.m^{-2}=RHS\)
\(LHS.m^{-2}=kgms^{-2}m^{-2}=kgm^{-1}s^{-2}\)
still \(m^{-2}\) missing!
What happened? Then comes \(\varepsilon_o\) with units of \(C^2N^{-1}m^{-2}\). Since \(\varepsilon_o\) would be in the denominator on the RHS, we have \(N.C^{-2}m^{2}\). This is not satisfactory.
Maybe the second division by \(4\pi r^2\) in the derivation for \(F\) along a radial line is not required?
We could have, energy density over a surface area of \(4\pi r^2\), \(\psi_A\) as
\(\psi_A=\psi.4\pi r^2\)
If \(\Delta E\) is the energy in the thickness of the thin spherical shell around the center particle of radius \(r\),
\(\Delta E=\psi_A.\Delta r\)
When this shell is infintely thin, \(\Delta r\to 0\),
\(\cfrac{d\,E}{d\,r}=\lim\limits_{\Delta r\to0}{\cfrac{\Delta E}{\Delta r}}=\psi_A\)
From which we obtain,
\(F_s=-\cfrac{d\,E}{d\,r}=-\psi_A\)
Strictly speaking we have lost the direction of \(F_s\) when we substituted in \(-\psi_A\) because
\(\psi_A\) cannot be negative,
\(|F_s|=\psi.4\pi r^2\)
But this force is distributed over the surface of the sphere of radius \(r\), as such the force along a radial line is,
\(|F|=\cfrac{|F_s|}{Area}=\cfrac{1}{4\pi r^2}\psi.4\pi r^2\)
\(|F|=\psi\)
If \(F\) is to have a inverse square law dependence then,
\(\psi=\cfrac{D}{r^2}\)
where \(D\) is a constant. Using \(E=mc^2\) we have
\(m=\cfrac{D}{c^2r^2}\)
but \(m\), mass density is in the time dimension so we have instead,
\(m=\cfrac{D}{c^2t^2}\)
which then suggests that \(m\) stretches out in time in a corresponding way as \(\psi\) stretches out in space. And instead of just a equivalence relationship we have a transform,
\(E(r)=m(t)c^2\)
from space (\(r\)) to time (\(t\)).
Simplicity rides again.
