Why A Positron And Deep Blue...

From the post "May the Hunt Begin" dated 1 Jun 2016,

$q=m.4\pi a^2_{\psi}= \cfrac { dq }{ dx }|_{x=a_{\psi}}$

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.