\(L=-\cfrac { 1 }{ 2 } \psi _{ c }\cfrac { \psi _{ c }+u }{ u } { e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }-\psi _{ n }\)
\(\cfrac{d}{dt}\left\{\cfrac{\partial\,L}{\partial\dot{x}}\right\}=\cfrac{d}{dt}\left\{\cfrac{1}{2}m2v\right\}=ma\)
since,
\(v^2=-\cfrac { 1 }{ m } \psi _{ c } \cfrac { \psi _{ c }+u }{ u }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }\)
And
\(\cfrac{\partial\,L}{\partial x}=-\cfrac{\partial\,\psi_n}{\partial x}=F_{\rho}\)
And we have,
\(F_{\rho}=ma\)
Wrong! This is not Newton's Law Of Motion. Which means \(\psi_n\) is not potential but potential density. In which case,
\(\cfrac{\partial\,L}{\partial x}=-\cfrac{\partial}{\partial x}\left\{\int_V\psi dV\right\}=\int_V{-\cfrac{\partial\,\psi}{\partial x}}dV=\int_V{F_{\rho}}dV=F\)
where \(F\) is the total force on the point particle and is in the direction of lesser \(\psi\) for a positive particle.
\(\cfrac{\partial\,L}{\partial x}=-\cfrac{\partial}{\partial x}\left\{\int_V\psi dV\right\}=\int_V{-\cfrac{\partial\,\psi}{\partial x}}dV=\int_V{F_{\rho}}dV=F\)
where \(F\) is the total force on the point particle and is in the direction of lesser \(\psi\) for a positive particle.
And we have,
\(F=ma\)
But, more importantly, should
\(\psi_n\) be \(\psi_n-\psi(x)\)
instead? In which case,
\(F=F_n-F(x)\)
where the resultant force \(F\) on the particle is \(F_n-F(x)\) in the direction of lesser \(\psi\) for a positive particle and reversed for a negative particle. \(F(x)\) being due to the containing particle.
Good Morning.
