Nature Has Orbits

From the post "Spin First Run Later" and "Photon Emission After Absorption" dated 15 Jul 2015, for $0\le x\lt x_z$

$\cfrac { d^{ 2 }x }{ dt^{ 2 } } =\cfrac { 1 }{ m }  \psi _{ c }  -{ \cfrac { u }{ \psi _{ c }+u }  }\cfrac { 1 }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } } (\cfrac { dx }{ dt } )^{ 2 }$

and for $x\gt x_z$, where $\psi(x_z)=\psi_{max}$,

$-\cfrac { d^{ 2 }x }{ dt^{ 2 } }=\cfrac { 1 }{ m }  \psi _{ c } +{ \cfrac { u }{ \psi _{ c }+u }  }\cfrac { 1 }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } } (\cfrac { dx }{ dt } )^{ 2 }$

Although,

$\cfrac { d^{ 2 }x }{ dt^{ 2 } } =0$ is trivial in most cases, it is not so here, especially in the latter equation where,

$0=\cfrac { 1 }{ m }  \psi _{ c } +{ \cfrac { u }{ \psi _{ c }+u }  }\cfrac { 1 }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } } (\cfrac { dx }{ dt } )^{ 2 }$

$(\cfrac { dx }{ dt } )^{ 2 }=v^2=-\cfrac { 1 }{ m }  \psi _{ c } \cfrac { \psi _{ c }+u   }{ u }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }$

which means $v=\cfrac { dx }{ dt }$ is complex and is perpendicular to $x$, the radial line joining the particle and the center of the $\psi$ cloud.

In that expression,

$\psi_c=\psi_{n}-\psi_{max}$

$u=ln(cosh(x-\cfrac { x_{ a } }{ 2 } ))$

which suggests that for each value of $x$ there is a corresponding value of $v$ by which the particle orbits about the center of the particle cloud at radius $r_{orbit}=x$.

The centrifugal force,

$F=m\cfrac{v^2}{x}$

This force is provided for by the difference in $\psi$ between the orbiting particle and the $\psi$ cloud.  This force points in the direction of higher $\psi$ towards $\psi_{max}$ at $x=x_z$.

$\psi_n-\psi(x)=F=m\cfrac{v^2}{x}$

which leads to,

$\psi _{ n }-\psi (x)=F=m\cfrac { v^{ 2 } }{ x } =-\cfrac { 1 }{ x } \psi _{ c }\cfrac { \psi _{ c }+u }{ u } { e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } }$

$\psi _{ n }\left\{ 1+\cfrac { 1 }{ x } \cfrac { \psi _{ c }+u }{ u } { e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } } \right\} =-ln(cosh(x-\cfrac {  x_{ a } }{ 2 })+\psi _{ max }\left\{ 1+\cfrac { 1 }{ x } \cfrac { \psi _{ c }+u }{ u } { e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } } \right\}$

$\left\{ \psi _{ n }-\psi _{ max }+ln(cosh(x-\cfrac { 1 }{ 2 } x_{ a }) \right\} \left\{ 1+\cfrac { 1 }{ x } \cfrac { \psi _{ c }+u }{ u } { e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } } \right\} =0$

which is true as,
$\require{cancel}$
$\cancelto{0}{\cfrac{d^2x}{dt^2}}+\cfrac{1}{m}\psi(x)=\cfrac{1}{m}\psi_{n}$

implies,

$\psi_{n}-\psi(x)=0$

$\psi _{ n }-\psi _{ max }+ln(cosh(x-\cfrac { 1 }{ 2 } x_{ a })=0$

We have just verified the possibility of $\cfrac { d^{ 2 }x }{ dt^{ 2 } } =0$ for $x\gt x_z$.  When $\cfrac { d^{ 2 }x }{ dt^{ 2 } } \gt 0$, $-\cfrac { d^{ 2 }x }{ dt^{ 2 } }$ points towards the origin and acts to provide for the centrifugal force.  $v^2$ is negative, $v$ is complex,  $iv$ is in the direction perpendicular to $x$.  The particle is in orbit around the origin.  Orbiting around the center of the particle $\psi$ cloud is the natural thing to do.

For the former case, $0\le x\lt x_z$, $\cfrac { d^{ 2 }x }{ dt^{ 2 } } =0$ just implies that the particle can instantaneously experience zero acceleration.

Note: First we verify the possibility of $v$ changing direction to $iv$ at $\cfrac{d^2x}{dt^2}=0$ then assert that when $\cfrac { d^{ 2 }x }{ dt^{ 2 } } \gt 0$, $-\cfrac { d^{ 2 }x }{ dt^{ 2 } }$ provides for the centrifugal force and the particle is thus in circular motion.  For $x\gt x_z$, where $\psi(x_z)=\psi_{max}$, the force on the particle is towards $\psi_{max}$, as such in the negative $x$ direction.