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