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