Spin First Run Later

From the post "No Solution But Exit Velocity Anyway" dated 14 Jun 2015,

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

where

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

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

A graph of the math components in the expression is,

We see that all components are positive, given

$\psi_c=\psi _{ n }-\psi_{max}\gt0$

$v_{max}$ is complex when

$\cfrac { 1 }{ m }  \psi _{ c }-\cfrac { d^{ 2 }x }{ dt^{ 2 } } \lt0$

$v_{max}=iv$

this velocity will be in the $ix$ direction; rotated from$+x$ by $\cfrac{\pi}{2}$, anticlockwise.

This can happen when the particle is accelerated further by other factors, a positive temperature particle , a charge, a $B$ field, etc.  The original differential equation will have an additive component that increases acceleration that is independent of $\psi_n$ but adds energy to the particle,

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

$\cfrac { d^{ 2 }x }{ dt^{ 2 } }+A_{other}$ is the total acceleration of the particle.  If on impact of a photon.

$\cfrac { 1 }{ m }  \psi _{ c }-\cfrac { d^{ 2 }x }{ dt^{ 2 } }-A_{other} \lt0$

The particle at zero velocity, first goes into a spin, with a velocity component perpendicular to the original direction $x$, then starts to accelerate along $x$ as the term $\cfrac { d^{ 2 }x }{ dt^{ 2 } }$ decreases.

In the case when the particle approaches light speed $c$ with a very large $A_{other}$ however, drag due to a space as a very light medium or drag due to entanglement that shares energy with the particle introduces a subtractive component to the original expression,

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

$D_{other}$ increases with increasing velocity such that at $v_{max}=c$,

$A_{other}=D_{other}$

and
$\require{cancel}$
$\cfrac { 1 }{ m }  \psi _{ c }-\cfrac { d^{ 2 }x }{ dt^{ 2 } }-\cancelto{0}{(A_{other}-D_{other}) }\lt0$

The particle spins at $c$ then accelerate along $x$.

Other energy input to the particle receiving a photon, results in spin until the expression,

$\cfrac { 1 }{ m }  \psi _{ c }-\cfrac { d^{ 2 }x }{ dt^{ 2 } }-A_{other} \lt0$

turns positive with decreasing $\cfrac { d^{ 2 }x }{ dt^{ 2 } }$.

(*) is a state equation, given $\cfrac{d^2x}{dt^2}$ and $\cfrac { 1 }{ m }  \psi _{ c }$, if  the expression, $\cfrac { 1 }{ m }  \psi _{ c }-\cfrac { d^{ 2 }x }{ dt^{ 2 } }$ is negative, velocity switches to the orthogonal direction.

This is very odd, when velocity and acceleration is perpendicular to each other, as odd as circular motion!

$\cfrac{d^2x}{dt^2}=\cfrac{v^2}{r}$

where $r$ is the radius of the spin from which we may obtain an expression for $r$,

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

Have a nice day.