No Solution But Exit Velocity Anyway

In the post "$\psi_{max}$ Is New And Important",

$\cfrac{d^2x}{dt^2}+\cfrac{1}{m}\psi(x)=\cfrac{1}{m}\psi_{n}$

In general,

$\psi(x)=-ln(cosh(x-\cfrac{1}{2}x_a)+C$

When $x=x_a$, $\psi(x_a)=0$,

$C=ln(cosh(\cfrac{1}{2}x_a))$

Since we consider the case when $x=0$, $\psi(x=0)=0$, it so happens that at $x=\cfrac{x_a}{2}$,

$C=\psi_{max}=ln(cosh(\cfrac{1}{2}x_a))$

So,

$\psi(x)=-ln(cosh(x-\cfrac{1}{2}x_a))+\psi_{max}$

and the differential equation becomes,

$\cfrac{d^2x}{dt^2}+\cfrac{1}{m}\left\{-ln(cosh(x-\cfrac{1}{2}x_a)+\psi_{max}\right\}=\cfrac{1}{m}\psi_{n}$ --- (*)

As $x$ changes with $t$,

$\cfrac { dx }{ dt } =\cfrac { 1 }{ m } \int {  \left\{ \psi _{ n }-\psi_{max} +ln(cosh(x-\cfrac { x_a } { 2 } ) \right\}  } dt$

$\cfrac { dx }{ dt } =\cfrac { 1 }{ m } \left\{ \psi _{ n }-\psi_{max}  \right\} .t+C+\cfrac { 1 }{ m } \int { ln(cosh(x-\cfrac { x_a } { 2 } ))dt }$

when $t=0$,

$\cfrac { dx }{ dt } |_{ t=0 }=0$

So, $C=0$

$\cfrac { dx }{ dt } =\cfrac { 1 }{ m } \left\{ \psi _{ n }-\psi_{max} \right\} .t+\cfrac { 1 }{ m } \int {ln(cosh(x-\cfrac { x_a } { 2 } ))dt }$

Integrating by $t$ again,

$x=\cfrac { 1 }{ 2m } \left\{ \psi _{ n }-\psi_{max}  \right\} .t^{ 2 }+\cfrac { 1 }{ m } \iint { ln(cosh(x-\cfrac { x_a } { 2 } ))dtdt }$

Consider,

$\cfrac { 1 }{ m } \iint { ln(cosh(x-\cfrac { x_a } { 2 } ))dtdt } =\cfrac { 1 }{ m } \iint { ln(cosh(x-\cfrac { x_a } { 2 } ))\cfrac { 1 }{ \cfrac { d^{ 2 }x }{ dt^{ 2 } }  } \cfrac { d^{ 2 }x }{ dt^{ 2 } } dtdt }$

But,

$\cfrac { d^{ 2 }x }{ dt^{ 2 } } dtdt=\cfrac { d }{ dt } \left( \cfrac { dx }{ dt }  \right) dtdt$

$d\left( \cfrac { dx }{ dt }  \right) dt=\cfrac { d(dx) }{ dt } dt=d(dx)$

So,

$\cfrac { d^{ 2 }x }{ dt^{ 2 } } dtdt=d(dx)$

and using (*),

$\cfrac { 1 }{ m } \iint { ln(cosh(x-\cfrac { x_a } { 2 } ))dtdt } =$

$\iint { ln(cosh(x-\cfrac { x_a } { 2 } ))\cfrac { 1 }{ \psi _{ n }+\left\{ ln(cosh(x-\cfrac { x_a } { 2 } )-\psi_{max}  \right\}  } d(dx) }$

$\iint { ln(cosh(x-\cfrac { x_a } { 2 } ))\cfrac { 1 }{ \psi _{ n }-\psi_{max} +ln(cosh(x-\cfrac { x_a } { 2 } ) } d(dx) }$

Consider,

$\cfrac { d }{ dx } \left( ln(cosh(x-\cfrac { x_a } { 2 } )) \right) =tanh(x-\cfrac { x_a } { 2 } )$

And we invert for the time being,

$\cfrac { dx }{ d\left( ln(cosh(x-\cfrac { x_a } { 2 } )) \right)  } =\cfrac { 1 }{ tanh(x-\cfrac { x_a } { 2 } ) }$

$\cfrac { d(dx) }{ dxd\left( ln(cosh(x-\cfrac { x_a } { 2 } )) \right)  } =\cfrac { d }{ dx } \left( \cfrac { 1 }{ tanh(x-\cfrac { x_a } { 2 } ) }  \right)$

$d(dx)=\cfrac { d }{ dx } \left( \cfrac { 1 }{ tanh(x-\cfrac { x_a } { 2 } ) }  \right) dxd\left( ln(cosh(x-\cfrac { x_a } { 2 } )) \right)$

If we let,

$L(x)=ln(cosh(x-\cfrac { x_a } { 2 } ))\cfrac { 1 }{ \psi _{ n }-\psi_{max} +ln(cosh(x-\cfrac { x_a } { 2 } ) )}$

So,

$\cfrac { 1 }{ m } \iint { ln(cosh(x-\cfrac { x_a } { 2 } ))dtdt }$

$= \iint { L(x)  } \cfrac { d }{ dx } \left( \cfrac { 1 }{ tanh(x-\cfrac { x_a } { 2 } ) }  \right) dxd\left( ln(cosh(x-\cfrac { x_a } { 2 } )) \right)$

$=-\iint { L(x)  } \cfrac { sech^{ 2 }(x-\cfrac { x_a } { 2 } ) }{ tanh^{ 2 }(x-\cfrac { x_a } { 2 } ) } dxd\left( ln(cosh(x-\cfrac { x_a } { 2 } )) \right)$

$=-\iint { L(x) } \cfrac { sech^{ 2 }(x-\cfrac { x_a } { 2 } ) }{ tanh^{ 2 }(x-\cfrac { x_a } { 2 } ) } { tanh(x-\cfrac { x_a } { 2 } ) }(dx)^{ 2 }$

$=-\iint { L(x)  } \cfrac { sech^{ 2 }(x-\cfrac { x_a } { 2 } ) }{ tanh(x-\cfrac { x_a } { 2 } ) } (dx)^{ 2 }$

$=-\iint { L(x) } \cfrac { 1 }{ cosh(x-\cfrac { x_a } { 2 } )sinh(x-\cfrac { x_a } { 2 } ) } (dx)^{ 2 }$

If we let,

$f(x)=cosh(x-\cfrac { x_a } { 2 } )$

and substitute back the expression for $L(x)$ then,

$\cfrac { 1 }{ m } \iint { ln(cosh(x-\cfrac { x_a } { 2 } ))dtdt }$

$=-\iint { ln(f(x))\cfrac { 1 }{ \psi _{ n }-\psi_{max}+ln(f(x) ) }   } \cfrac { 1 }{ f(x)f^{'}(x)} (dx)^{ 2 }$

If we let,

$u=ln(f(x))$

$\cfrac { du }{ dx } =\cfrac { 1 }{ f(x) } f^{ ' }(x)$

and,

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

then,

$\cfrac { 1 }{ m } \iint { ln(cosh(x-\cfrac { x_a } { 2 } ))dtdt }$

$= -\iint { \cfrac { u }{ \psi _{ c }+u }  } \cfrac { 1 }{ (f^{ ' }(x))^{ 2 } } \cfrac { du }{ dx } (dx)^{ 2 }$

Since,

$e^{ u }=f(x)$

$\cfrac { de^{ u } }{ dx } =e^{ u }\cfrac { du }{ dx } =f^{ ' }(x)$

We have,

$-\iint { \cfrac { u }{ \psi _{ c }+u }  } e^{ -2u }\left( \cfrac { dx }{ du }  \right) ^{ 2 }\cfrac { du }{ dx } (dx)^{ 2 }$

$= -\iint { \cfrac { u }{ \psi _{ c }+u }  } e^{ -2u }\cfrac { dx }{ du } (dx)^{ 2 }$

$=-\iint { \cfrac { u }{ \psi _{ c }+u }  } e^{ -2u }\left( \cfrac { dx }{ du }  \right) ^{ 3 }(du)^{ 2 }$

Since,

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

$\cfrac { du }{ dx } =tanh(x-\cfrac { x_{ a } }{ 2 } )$

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

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

$e^{ 2u }-1=sinh^{ 2 }(x-\cfrac { x_{ a } }{ 2 } )$

$(\cfrac { du }{ dx } )^{ 2 }=\cfrac { e^{ 2u }-1 }{ e^{ 2u } }$

$(\cfrac { du }{ dx } )^{ 3 }=\left( \cfrac { e^{ 2u }-1 }{ e^{ 2u } }  \right) ^{ 3/2 }$

Finally,

$\cfrac { 1 }{ m } \iint { ln(cosh(x-\cfrac { x_a } { 2 } ))dtdt }$

$= -\iint { \cfrac { u }{ \psi _{ c }+u }  } \cfrac { e^{ u } }{ \left( { e^{ 2u }-1 } \right) ^{ 3/2 } } (du)^{ 2 }$

And so,

$x=\cfrac { 1 }{ 2m } \left\{ \psi _{ n }-\psi_{max}  \right\} .t^{ 2 }+\cfrac { 1 }{ m } \iint { ln(cosh(x-\cfrac { x_a } { 2 } ))dtdt }$

$x=\cfrac { 1 }{ 2m } \psi _{ c } .t^{ 2 } -\iint { \cfrac { u }{ \psi _{ c }+u }  } \cfrac { e^{ u } }{ \left( { e^{ 2u }-1 } \right) ^{ 3/2 } } (du)^{ 2 }$

The double integral cannot be solved easily.  If we retrace and consider,

$x=\cfrac { 1 }{ 2m }\psi _{ c } .t^{ 2 } -\iint { \cfrac { u }{ \psi _{ c }+u }  } \cfrac { e^{ u } }{ \left( { e^{ 2u }-1 } \right) ^{ 3/2 } } \cfrac{du}{dt}dudt$

$x=\cfrac { 1 }{ 2m } \psi _{ c }  .t^{ 2 }-\iint { \cfrac { u }{ \psi _{ c }+u }  } \cfrac { e^{ u } }{ \left( { e^{ 2u }-1 } \right) ^{ 3/2 } } \cfrac{du}{dx}.\cfrac{dx}{dt}dudt$

$\cfrac{dx}{dt}=\cfrac { 1 }{ m } \psi _{ c }  .t-\int { \cfrac { u }{ \psi _{ c }+u }  } \cfrac { e^{ u } }{ \left( { e^{ 2u }-1 } \right) ^{ 3/2 } } \cfrac{du}{dx}.\cfrac{dx}{dt}du$

Using,

$\cfrac { du }{ dx } =(\cfrac { e^{ 2u }-1 }{ e^{ 2u } } )^{1/2}$

$\cfrac{dx}{dt}=\cfrac { 1 }{ m }  \psi _{ c }   .t-\int { \cfrac { u }{ \psi _{ c }+u }  } \cfrac { e^{ u } }{ \left( { e^{ 2u }-1 } \right) ^{ 3/2 } } (\cfrac { e^{ 2u }-1 }{ e^{ 2u } } )^{1/2}.\cfrac{dx}{dt}du$

$\cfrac{dx}{dt}=\cfrac { 1 }{ m } \psi _{ c }  .t-\int { \cfrac { u }{ \psi _{ c }+u }  } \cfrac { 1 }{ \left( { e^{ 2u }-1 } \right)  }.\cfrac{dx}{dt}du$

Differentiating wrt $u$,

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

Multiplying by $\cfrac { du }{ dt }$,

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

Since,

$\cfrac { du }{ dx } \cfrac { dx }{ dt }=\cfrac { du }{ dt }  =(\cfrac { e^{ 2u }-1 }{ e^{ 2u } } )^{ 1/2 }\cfrac { dx }{ dt }$

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

where,

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

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

Expression (**)  does not solve for the differential equation (*), but relates the acceleration of the particle with its velocity.  We know that the particle is oscillating, its velocity is maximum just before it experiences a retarding force at the center of oscillation, ie,

$\cfrac { d^{ 2 }x }{ dt^{ 2 } } =0$

This is the point at which acceleration turn from positive to negative (or vice versa).

$\cfrac { dx }{ dt }=v_{max}$

Just before the particle experience deceleration, its velocity is maximum.

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

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

When this occurs at $x=x_a$, the particle is ejected,  (the center of oscillation need not be the center of $\psi$), $\psi$ is spherical centered at $x=0$.)

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

As,   $\psi_c=\psi_n-\psi_{max}=\psi_n-ln(cosh(\cfrac { x_{ a } }{ 2 } ))$

$v^{ 2 }_{ { max } }=\cfrac { 1 }{ m } \cfrac { \psi _{ n } }{ \psi _{ max } }\left\{ \psi _{ n }-\psi _{ max } \right\}  e^{ \psi _{ max } }\left( { e^{ 2\psi _{ max } }-1 } \right) ^{ 1/2 }$

where $m$ is the mass density of the particle.  The unit dimension of this expression for $v^2_{max}$ is consistent.  $\psi_{max}$ is the constraint presented by the non-transit particle or the bulk of the particle cloud,  $\psi_n$ is the energy density state of the transit particle after absorbing a photon,  

$\psi_n=\psi_o+\psi_{photon}$

$\psi_o$ is the initial energy state of the transit particle before its encounter with the photon, $\psi_{photon}$.  This expression is valid for $\psi_n-\psi_{max}\gt0$; only when the particle gains more energy than the energy barrier presented by the non-transit particle, $\psi_n\gt\psi_{max}$ is $v_{max}$ non zero.

Although all parameters that might affect $v_{max}$ are here and the unit dimension checks out, this expression for $v_{max}$ is at best reasonable and needs to be checked.

Another way to look at the situation when the particle is ejected is to see that, as long as $v_{max}$ is positive at $x=x_a$, no matter under acceleration or deceleration the particle will be at $x=x_a+\delta x$ in the next instant and so, be ejected from the cloud.  From expression (**), if the particle is accelerating at $x=x_a$ with some positive $v_{max}$, the particle is certain to be ejected.  If the particle is decelerating,

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

In the expression (**) for $x=x_a$,

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

deceleration adds to the value of $v$.  The least  contribution it makes to $v=\cfrac { dx }{ dt }$ is when it is zero. So,

$\cfrac { d^{ 2 }x }{ dt^{ 2 } }=0$

is the limiting condition to eject the transit particle.

Thank you very much.  'Til tomorrow!