From the previous post "Equating To A Indefinite Integral" dated 10 Jun 2016,
\(\dot { x } ^{ 3 }=-\cfrac { 1 }{ 4\pi } \cfrac { q }{ m } a_{ \psi }\)
where the negative sign signifies an attractive force.
\( \dot { x } =-\sqrt [ 3 ]{ \cfrac { 1 }{ 4\pi } \cfrac { q }{ m } a_{ \psi } }\)
\( \cfrac { \partial \, T }{ \partial \, x } =m\dot { x } \cfrac { \partial \, \dot { x } }{ \partial \, x } \)
\( \cfrac { \partial \, T }{ \partial \, x } =-\sqrt [ 3 ]{ \cfrac { q }{ 4\pi } a_{ \psi } } .\cfrac { \partial \, \dot { x } }{ \partial \, x }.m^{ 2/3 }\)
Only a decreasing \(\dot{x}\) along \(x\) is,
\(\cfrac { \partial \, T }{ \partial \, x }\gt 0\)
that would make,
\(q_{ a_{ \psi } } =3k-2\cfrac { \partial \, T }{ \partial \, x }|_{a_{\psi}}\lt0\)
where
\(\cfrac{\partial\psi}{\partial x}=\cfrac{\partial V}{\partial x}+\cfrac{\partial T}{\partial x}=k\)
If \(\dot{x}\) increases along \(x\) or decreases less along \(x\),
\(\cfrac { \partial \, T }{ \partial \, x }\lt 0\)
or
\(\cfrac { \partial \, T }{ \partial \, x }_1\lt \cfrac { \partial \, T }{ \partial \, x }_2\)
such that,
\(q_{ a_{ \psi } } =3k-2\cfrac { \partial \, T }{ \partial \, x }|_{a_{\psi}}\gt0\)
the particle switches sign.
This seems to mark the boundary when two particles with interacting \(\psi\) turn from attractive (interacting as waves) to repulsive (interacting) as particles. It is not. The particles' \(\psi\)s are already interacting and they are interacting as waves. After the sign reversal, the particles remains together.
\(3k=2\cfrac { \partial \, T }{ \partial \, x }|_{a_{\psi}}\)
and is neutral (neither attractive nor repulsive) to each other. We can simplify to obtain,
\(3 \cfrac { \partial V\, }{ \partial \, x }|_{a_{\psi}} =-\cfrac { \partial \, T }{ \partial \, x }|_{a_{\psi}} \)
this is valid only at the neutral point as \( \cfrac { \partial V\, }{ \partial \, x }\) changes with \(\cfrac { \partial \, T }{ \partial \, x } \) along \(x\). \(k\) given \(\psi\) is a constant.
It is not likely that \(\dot{x}\) increases. In the case when \(\dot{x}\) decreases less along \(x\), \(\psi\) spreads itself away from the center of the particle as it reduces speed along \(x\). In free space, a particle is negative, in a medium where \(\psi\) is encouraged to spread, the particle is positive. eg. an electron in free space and an electron on glass (\(Si\)), respectively.
Axiomatic but a good guess.
