Leaving Without Me?

The photon is emitted before the particle escapes (if it escapes) as,

$|\cfrac { d^{ 2 }x }{ dt^{ 2 } }|-\cfrac { 1 }{ m_{photon} }  \psi _{ c }\lt |\cfrac { d^{ 2 }x }{ dt^{ 2 } }|-\cfrac { 1 }{ m }  \psi _{ c }$

The photon escapes the particle tangentially and the particle escape $\psi$  tangentially.  Both will be in a spin of radius, if they escape,

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

The ratio of energies,

 $\cfrac {  \psi _{ c } }{ m v^2}$

where

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

$\psi_n=\psi _{ photon }+\psi_{o}$

$\psi_n$ is the elevated energy level of the particle from its ground state $\psi_o$ after receiving the photon $\psi_{photon}$.  $\psi_{max}$ is the maximum $\psi$ of the confining particle cloud.

is most interesting.  Given a non transit (confining) particle, the term,

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

is fixed.  It is also at maximum at the boundary of $\psi(x)$. (To be proven...)  The energy ratio, $\cfrac {  \psi _{ c } }{ m v^2}$ determines the radius of the escape spin.

This is the second way, a particle goes into a spin.  The first being on head on retardation by a temperature particle that stops the particle before collision, from the post "No Temperature Particle At All, Zero" date 12 Jul 2015.

Note:

Both transit and non-transit particles are of the same type, otherwise their energy densities would not interact.  The non-transit particle is considered as an extended spherical $\psi$ cloud around the particle center. It could be the superposition of many particles of the same type.  The transit particle is considered as a point particle inside $\psi$ of the non transit particle.