Photon Emission After Absorption

From the post "$\psi_{max}$ Is New And Important" dated 13 Jul 2015,

$F=\psi_{n}-\psi(x)=m\cfrac{dv}{dt}=m\cfrac{d^2x}{dt^2}$

In writing down this equation,  $\psi(x)$ of the non transit particle around the transit particle was taken as reference to give an expression for the force on the transit particle.  If the energy density, $\psi_n$ of the transit particle is higher than its surrounding, a positive resultant is obtained.  This positive resultant points towards the positive direction of $x$.  This direction however may not be the right direction.

When $\psi_n\gt\psi(x)$, the force on the transit particle is less than the force on the non-transit particle.  Both particles are free to move.  But in what direction?  The expression,

$|F|=\psi$

from the post "Opps Lucky Me" dated 25 May 2015, $F$ the Newtonian force has lost its direction.

In the case above, the transit particle itself having higher $\psi$ will move towards region of higher $\psi$ in the non-transit particle.  The non-transit particle will move relatively in the opposite direction.  If the transit particle has comparatively lower $\psi$, it will move towards region of lower $\psi$ in the non-transit particle.  In both scenarios, the force on the particle reduces to zero.  When the $\psi$s of both particles are equal there is no force; $F=0$.

This avoids a run away scenario where the particle move into a region of $\psi$ that results in a greater resultant force which than run the particle further into the region that results in a even greater force.  The force on the particle should result in motion in a direction that bring about a reduction of the force itself.

The above equation does not capture this information.

$-F=\psi_{n}-\psi(x)=-m\cfrac{dv}{dt}=-m\cfrac{d^2x}{dt^2}$

is equally valid, for the situation where the direction for reducing $F$ is negative.  This occurs when $x\gt \cfrac{x_a}{2}$.  The graph below shows the situation when $\psi_n\gt\psi(x)$ at $x=x_v$ and $x=x_c$.

The change in sign of $F$ and so $\cfrac{d^2x}{dt^2}$ at $x=x_c$ leads to,

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

($\cfrac { d^{ 2 }x }{ dt^{ 2 } }$ was substituted into the derivation again, which switches the sign of the second term on the RHS.)

then

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

Since $\cfrac { d^{ 2 }x }{ dt^{ 2 } }\lt0$,

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

This second expression for $v$ and $\cfrac { d^{ 2 }x }{ dt^{ 2 } }$ shed light on the fate of the photon absorbed.  As developed in the post "Spin First Run Later", when the LHS of this expression is negative, the velocity of the particle $v=\cfrac{dx}{dt}$, becomes complex, $iv$.  This signifies that the particle's velocity has switched to a perpendicular direction $ix$.

Mean while...

The photon that was absorbed remains with the transit particle.  If it retain its entity, it is also subjected to the same equation (*) above, but,

$m_{photon}\lt m$

the photon mass density is very much less than the particle mass density, as such,

$\cfrac { 1 }{ m_{photon} }  \psi _{ c }\gt \cfrac { 1 }{ m }  \psi _{ c }$

and so,

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

The velocity of the photon will switch direction before the particle.  When this happens the photon is emitted in a direction perpendicular to the direction of travel of the particle.  This happens when the particle returns from beyond maximum $\psi$, at the far side, $x=x_c$ when the above expression (*) applies.  This does not happen at $x=x_v$.  On this side, acceleration starts at a maximum value and reduces as velocity increases.  Under some circumstances when a switch do occurs, the particle will switch first because of its higher mass density (note the changed position of the terms involved) and carry the photon with it.  (There is a mistake in the post "$\psi_{max}$ Is New And Important" dated 13 Jul 2015, after the photon is emitted the particle does not return to it original position and photon emission occurs when the force $F$ is negative, only on the side $x=x_c$)

The particle and the photon after being absorbed move as a whole with total mass density, $m_n$,

$m_n=m+m_{photon}$

with a common velocity and acceleration.  Confining forces within the particle holds the photon such that both experiences the same acceleration.  For the photon,

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

where $F_p$ from the particle reduces the acceleration of the photon to that of the more massive particle.  This force, $F_p$ is along the $-x$ direction, velocity along $-ix$ is not affected by this force.  The photon will escape the particle with velocity, $-iv$, when the switch occurs when,

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

This is how and when the absorbed photon is emitted.