Into A Pile Of Deep Shit

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 $\cfrac{1}{\gamma^2}=\left( 1-\cfrac { \dot { x } ^{ 2 } }{ c^{ 2 } }  \right)$

Together with,

$\psi =T+V$

we have,

$\left( 1+\cfrac { 1 }{ \gamma ^{ 2 } }  \right) \cfrac { \partial \, \psi  }{ \partial \, t_{ c } } =2\left( \cfrac { \partial \, \psi  }{ \partial \, t_{ c } } -\cfrac { \partial \, T }{ \partial \, t_{ c } }  \right)$

$\left( 1-\cfrac { 1 }{ \gamma ^{ 2 } }  \right) \cfrac { \partial \, \psi  }{ \partial \, t_{ c } } =2\cfrac { \partial \, T }{ \partial \, t_{ c } }$

$\left( 1-\cfrac { 1 }{ \gamma ^{ 2 } }  \right) \left(\cfrac { \partial \, \psi  }{ \partial \, t_{ c } } + \cfrac { \partial \, \psi  }{ \partial \, x }\cfrac { \partial \, x }{ \partial \, t_{ c } }\right)=2\left(\cfrac { \partial \, T  }{ \partial \, t_{ c } } + \cfrac { \partial \, T  }{ \partial \, x }\cfrac { \partial \, x }{ \partial \, t_{ c } }\right)$

For $\cfrac { \partial \, x }{ \partial \, t_{ c } } \ne 0$,  and $\cfrac { \partial \, \psi  }{ \partial \, t_{ c } }=0$ as $\psi$ is stationary, $\left.\psi\right|_{x}=k$ given $x$,

$\left( 1-\cfrac { 1 }{ \gamma ^{ 2 } }  \right) \cfrac { \partial \, \psi  }{ \partial \, x }=2\left(\cfrac { \partial \, T  }{ \partial \, t_{ c } } + \cfrac { \partial \, T  }{ \partial \, x }\cfrac { \partial \, x }{ \partial \, t_{ c } }\right) \cfrac { \partial \, t_{ c } }{ \partial \, x }$

Here, given a position $x$, both $\psi$ do not change through time $t_c$ but changes along $x$ as we pass by with $\dot{x}=\cfrac { \partial \, x }{ \partial \, t_{ c } }$ through fields of stationary values of $\psi$.

$\left( 1-\cfrac { 1 }{ \gamma ^{ 2 } }  \right) \cfrac { \partial \, \psi  }{ \partial \, x } =4\cfrac { \partial \, T }{ \partial \, x }$

Similarly,

$\left( 1+\cfrac { 1 }{ \gamma ^{ 2 } }  \right) \cfrac { \partial \, \psi  }{ \partial \, x } =4\cfrac { \partial V\,  }{ \partial \, x }$

With these we take a good hard look at,

$\cfrac { dq }{ dx } =\left( \cfrac { 3\gamma ^{ 2 }+1 }{ \gamma ^{ 2 }+1 }  \right) \cfrac { \partial V\,  }{ \partial \, x } +\cfrac { \partial \, T }{ \partial \, x }$

where $2\left( 1+\cfrac { 1 }{ \gamma ^{ 2 } }  \right) ^{ -1 }+1=\left( \cfrac { 3\gamma ^{ 2 }+1 }{ \gamma ^{ 2 }+1 }  \right)$

$\cfrac { dq }{ dx } =\left( \cfrac { 3\gamma ^{ 2 }+1 }{ \gamma ^{ 2 }+1 }  \right) .\cfrac { 1  }{ 4 } \left( 1+\cfrac { 1 }{ \gamma ^{ 2 } }  \right) \cfrac { \partial \, \psi  }{ \partial \, x } +\cfrac { 1  }{ 4 } \left( 1-\cfrac { 1 }{ \gamma ^{ 2 } }  \right) \cfrac { \partial \, \psi  }{ \partial \, x }$

$\cfrac { dq }{ dx } =\left( 2\left( 1+\cfrac { 1 }{ \gamma ^{ 2 } }  \right) ^{ -1 }+1 \right) .\cfrac { 1  }{ 4 } \left( 1+\cfrac { 1 }{ \gamma ^{ 2 } }  \right) \cfrac { \partial \, \psi  }{ \partial \, x } +\cfrac { 1 }{ 4 } \left( 1-\cfrac { 1 }{ \gamma ^{ 2 } }  \right) \cfrac { \partial \, \psi  }{ \partial \, x }$

$\cfrac { dq }{ dx } = \left( \cfrac{1}{2}+\cfrac { 1 }{ 4} \left( 1+\cfrac { 1 }{ \gamma ^{ 2 } }  \right) +\cfrac { 1 }{ 4 } \left( 1-\cfrac { 1 }{ \gamma ^{ 2 } }  \right)  \right) .\cfrac { \partial \, \psi  }{ \partial \, x }$

$\cfrac { dq }{ dx } =\cfrac { \partial \, \psi  }{ \partial \, x } = \cfrac { d\psi }{ dx }$

as $\psi$ is independent of $t_c$ given $x$,

$q=\psi$

which is the reason we needed in the post "Why A Positron And Deep Blue..." dated 01 Jun 2016, for the expression,

$\cfrac{dq}{dx}=\cfrac{\partial\,\psi}{\partial\,t}\cfrac{\partial\,t}{\partial\,x}+\cfrac{\partial\,\psi}{\partial\,x}$

and we don't have problem with this.