Which get us to the difficult part, if ONLY \(g\) particles interacts with the gravitational field, what is \(m_e\) associated with a \(e^{-}\) particle? A
(\(g^{+}\), \(T^{+}\), \(e^{-}\))
particle would be as heavy as a neutron! Strictly speaking, an electron can only be moved by an electric force in an electric field and not by any physical force,
\(F\ne m_ea\)
But consider,
\(F_E=\cfrac{q_1q_2}{4\pi \varepsilon_o r^2}\) and,
\(V_E=\cfrac{q_1q_2}{4\pi \varepsilon_o r}\)
\(\cfrac{d\,V_E}{dt}=\cfrac{q_1q_2}{4\pi \varepsilon_o r^2}.-1.\cfrac{d\,r}{dt}=F_E.-1.\cfrac{d\,r}{dt}\)
The minus sign indicates that potential energy decreases in the direction of increasing \(r\). But the Newtonian force and the associated power is,
\(P=F.v=ma.v\) we have also assume that mass is time invariant here
So,
\(ma.v=F_E.\cfrac{d\,r}{dt}\)
where along a field line, \(v=\cfrac{d\,r}{dt}\), it is then possible that,
\(F_E=ma\) --- (*)
that is to say, in an electric field, it is possible to associate the product of a mass and an acceleration with the electric force, if we let \(v=\cfrac{d\,r}{dt}\) to be along a field line. This however is valid only in an electric field, where \(r\) is valid and we are moving along \(r\) a electric field line. Devoid of an electric field expression (*) is not valid.
Consider now,
\(\cfrac{d\,F_E}{dt}=\cfrac{q_1q_2}{4\pi \varepsilon_o r^3}.-2.\cfrac{d\,r}{dt}=F_E.-2.\cfrac{1}{r}\cfrac{d\,r}{dt}=F_E.-2.\cfrac{d\,ln(r)}{d\,t}\)
\(\cfrac{d\,F}{dt}=m\cfrac{d\,a}{dt}+a\cfrac{d\,m}{dt}\)
If we asume that \(\cfrac{d\,m}{dt}=0\), then,
\(m\cfrac{d\,a}{dt}=-2F_E\cfrac{d\,ln(r)}{d\,t}=ma\cfrac{d\{-2ln(r)\}}{d\,t}\)
\(\cfrac{d\,ln(a)}{dt}=\cfrac{d\,ln(r^{-2})}{d\,t}\)
\(ln(a)=ln(r^{-2})+lnA\) where \(ln(A)\) is constant of integration,
\(a=A.\cfrac{1}{r^2}\)
If \(A=\cfrac{1}{m}\cfrac{q_1q_2}{4\pi\varepsilon_o}\)
we once again obtain a consistent expression for \(F_E\) and so \(V_E\). In arriving at this expression, we have assumed,
\(\cfrac{d\,m}{dt}=0\)
that this, the associated mass \(m\), is time invariant. So, calculations for \(r\) along a field line using,
\(F_E=ma\)
\(m\) being a constant will give us the consistent answers for \(r\) and \(V_E\), given \(F_E\).
As such, it is possible to associate masses, \(m_e\) and \(m_p\) with the particles \(e^{-}\) and \(p^{+}\) respectively, but only in an electric field and all motions are restricted to be along the field lines.
\(wt=m_e.g\)
the weight of an electron, is strictly absurd.
