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.
