The expression for \(\psi\) from the previous post "Not A Wave, But Work Done!",
\(\ddot { x } \cfrac { \partial \, \psi }{ \partial \, t } =-\cfrac{c^3}{\sqrt{2}}\cfrac { \partial ^{ 2 }\psi }{ \partial \, x^{ 2 } }e^{-i\pi/4}\)
indicates a phase lag between a change in \(\psi\) with time and its second derivative change in \(x\).
This means that we have to apply an increasing force, \(F\)
\(-\cfrac{\partial\,\psi}{\partial\,x}=F\)
such that,
\(-\cfrac{\partial^2\,\psi}{\partial\,x^2}=\cfrac{\partial\,F}{\partial\,x}\ne 0\)
in order for a change in energy \(\psi\). The negative sign shows that the system opposes the force applied and work is done to the system and energy transferred to the particle. This change in energy lags the changing force by \({\pi}/{4}\), and there is a delay in the effect of this changing force.
\(\ddot { x } \cfrac { \partial \, \psi }{ \partial \, t } =-\cfrac{c^3}{\sqrt{2}}(-\cfrac{\partial\,F}{\partial\,x})e^{-i\pi/4}\)
\(\ddot { x } \cfrac { \partial \, \psi }{ \partial \, t } =\cfrac{c^3}{\sqrt{2}}(\cfrac{\partial\,F}{\partial\,x})e^{-i\pi/4}\)
This means for a gain in \(\psi\) (either \(T\), kinetic energy or \(V\), potential energy or both), \(F\) must be increasing and the effects of \(F\) is delayed in time. Inertia!
