\(q=m.4\pi a^2_{\psi}= \cfrac { dq }{ dx }|_{x=a_{\psi}}\)
Given an example of a particle with a surrounding field,
\(a_{\psi}\) is along \(x\) in the diagram. So,
\(\cfrac{dq}{dx}=\cfrac{\partial\,\psi}{\partial\,t}\cfrac{dt}{dx}+\cfrac{\partial\,\psi}{\partial\,x}\)
since there is no other entity in this model other than \(\psi\), which is a big assumption here.
\(\cfrac{dq}{dx}=\cfrac{1}{c}\cfrac{\partial\,\psi}{\partial\,t}+\cfrac{\partial\,\psi}{\partial\,x}\)
when the particle exist along \(t_c\),
\(\cfrac{dq}{dx}=\cfrac{1}{c}\cfrac{\partial\,\psi}{\partial\,t_c}+\cfrac{\partial\,\psi}{\partial\,x}\)
From the post "My Own Wave Equation" dated 20 Nov 2014,
\(\left(1+\cfrac{1}{\gamma^2} \right)\cfrac { \partial \, \psi }{ \partial \, t_{ c } }=2\cfrac { \partial V\, }{ \partial \, t_{ c } }\)
where \(V\) is the potential energy of the particle and \(\cfrac{1}{\gamma^2}=\left( 1-\cfrac { \dot { x } ^{ 2 } }{ c^{ 2 } } \right)\)
We have,
\(\cfrac{dq}{dx}=\cfrac{2}{c}\left(1+\cfrac{1}{\gamma^2} \right)^{-1}.\cfrac { \partial V\, }{ \partial \, t_{ c } }+\cfrac{\partial\,\psi}{\partial\,x}\)
Assuming that \(V\) is stationary in space and changes only because of the particle is at velocity \(c\),
\(\cfrac { dq }{ dx } =\cfrac { 2 }{ c } \left( 1+\cfrac { 1 }{ \gamma ^{ 2 } } \right) ^{ -1 }.\cfrac { \partial V\, }{ \partial \, x } \cfrac { \partial x }{ \partial \, t_{ c } } +\cfrac { \partial \, \psi }{ \partial \, x } \)
\( \cfrac { dq }{ dx } =2\left( 1+\cfrac { 1 }{ \gamma ^{ 2 } } \right) ^{ -1 }.\cfrac { \partial V\, }{ \partial \, x } +\cfrac { \partial \, \psi }{ \partial \, x } \)
Since, \(\psi =T+V\) is the total energy, \(T\) is the kinetic energy,
\( \cfrac { dq }{ dx } =\left( \cfrac { 3\gamma ^{ 2 }+1 }{ \gamma ^{ 2 }+1 } \right) \cfrac { \partial V\, }{ \partial \, x } +\cfrac { \partial \, T }{ \partial \, x } \)
\(\cfrac{1}{\gamma^2}\rightarrow 0\), \(\dot{x}=c\)
\( \cfrac { dq }{ dx } =3 \cfrac { \partial V\, }{ \partial \, x } +\cfrac { \partial \, T }{ \partial \, x } \)
\({q}=\left[3 \cfrac { \partial V\, }{ \partial \, x } +\cfrac { \partial \, T }{ \partial \, x } \right]_{x=a_{\psi}}.a_{\psi}\)
\(\psi\) is a wave in the \(ix\) direction, both \(\cfrac { \partial\,V }{ \partial \, x }\) and \(\cfrac { \partial\, T }{ \partial \, x }\) are along the radial line, in the \(x\) direction.
\(\cfrac { \partial\,V }{ \partial \, x }\) represents a force inside the particle,
\(F_{n}=\cfrac { \partial\,V }{ \partial \, x }\)
that pulls \(\psi\) to the particle center \(x=0\), when \(V\) decreases with \(x\). When the total \(\psi\) along \(x\) is conserved then, at \(x=a_{\psi}\),
\({q}\gt0\)
when
\(3\cfrac { \partial\,V }{ \partial \, x }|_{a_{\psi}}\gt-\cfrac { \partial\,T }{ \partial \, x }|_{a_{\psi}}\)
and,
\({q}\lt0\)
when
\(3\cfrac { \partial\,V }{ \partial \, x }|_{a_{\psi}}\lt-\cfrac { \partial\,T }{ \partial \, x }|_{a_{\psi}}\)
Without an external force, \(\cfrac { \partial\,V }{ \partial \, x }|_{a_{\psi}} \) and \(\cfrac { \partial\,T }{ \partial \, x }|_{a_{\psi}}\) are expected to change in opposite direction, but,
\(\cfrac { \partial\,\psi }{ \partial \, x }|_{a_{\psi}}\ne0\)
so,
\(\cfrac { \partial\,V }{ \partial \, x }|_{a_{\psi}}\ne-\cfrac { \partial\,T }{ \partial \, x }|_{a_{\psi}}\)
as \(\psi=V+T\).
The sign of the particle can change! Either there is a positive electron charge that in spin produces a \(B\) or \(T\) field or positive and negative particles are not differentiated by their oscillating energies.
The former is likely because a positron is known to exist, but borne here, it is not an antimatter.
And of the six particles in total, now we have eighteen, because
\(3\cfrac { \partial\,V }{ \partial \, x }|_{a_{\psi}}=-\cfrac { \partial\,T }{ \partial \, x }|_{a_{\psi}}\)
\({q}=0\)
creates a charge neutral particle and makes three possibilities in total for each particle.
This requires a rethink...in deep blue.
