Friday, July 17, 2015

Reach, Touch, A...

If we allow for the second constant of integration in \(F\), then

\(\psi (x)=-ln(cosh(x-\cfrac { 1 }{ 2 } x_{ a }))+Ax+\psi _{ max }\)

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

From,

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

\( \cfrac { d^{ 2 }x }{ dt^{ 2 } } +\cfrac { 1 }{ m } \left\{ -ln(cosh(x-\cfrac { 1 }{ 2 } x_{ a }))+Ax+\psi _{ max } \right\} =\cfrac { 1 }{ m } \psi _{ n }\)

\( \cfrac { d^{ 2 }x }{ dt^{ 2 } } =\cfrac { 1 }{ m } \left\{ \psi _{ n }-\psi _{ max }+ln(cosh(x-\cfrac { 1 }{ 2 } x_{ a }))-Ax \right\} \)

\( \psi _{ c }=\psi _{ n }-\psi _{ max }\)

\( \cfrac { d^{ 2 }x }{ dt^{ 2 } } =\cfrac { 1 }{ m } \left\{ \psi _{ c }+ln(cosh(x-\cfrac { 1 }{ 2 } x_{ a }))-Ax \right\} \)

\( \cfrac { d^{ 2 }x }{ dt^{ 2 } } =\cfrac { 1 }{ 2m } \psi _{ c }t^{ 2 }+\cfrac { 1 }{ m } \iint { \left\{ ln(cosh(x-\cfrac { 1 }{ 2 } x_{ a }))-Ax \right\}  } dtdt\)

Let,

\( u=ln(cosh(x-\cfrac { 1 }{ 2 } x_{ a }))-Ax\)

\( \cfrac { d^{ 2 }x }{ dt^{ 2 } } =\cfrac { 1 }{ 2m } \psi _{ c }t^{ 2 }+\cfrac { 1 }{ m } \iint { u } \cfrac { 1 }{ \cfrac { d^{ 2 }x }{ dt^{ 2 } }  } \cfrac { d^{ 2 }x }{ dt^{ 2 } } dtdt\)

\( \cfrac { d^{ 2 }x }{ dt^{ 2 } } =\cfrac { 1 }{ 2m } \psi _{ c }t^{ 2 }+\iint { \cfrac { u }{ \psi _{ { c } }+u }  } d(dx)\)

\( \cfrac { du }{ dx } =tanh(x-\cfrac { 1 }{ 2 } x_{ a })-A\)

\( \cfrac { dx }{ du } =\cfrac { 1 }{ tanh(x-\cfrac { 1 }{ 2 } x_{ a })-A } \)

\( d(dx)=\cfrac { 1 }{ tanh(x-\cfrac { 1 }{ 2 } x_{ a })-A } dudu\)

\( \cfrac { d^{ 2 }x }{ dt^{ 2 } } =\cfrac { 1 }{ 2m } \psi _{ c }t^{ 2 }+\iint { \cfrac { u }{ \psi _{ { c } }+u } \cfrac { 1 }{ tanh(x-\cfrac { 1 }{ 2 } x_{ a })-A }  } dudu\)

\( \cfrac { d^{ 2 }x }{ dt^{ 2 } } =\cfrac { 1 }{ 2m } \psi _{ c }t^{ 2 }+\iint { \cfrac { u }{ \psi _{ { c } }+u } \cfrac { 1 }{ tanh(x-\cfrac { 1 }{ 2 } x_{ a })-A }  } \cfrac { du }{ dt } dudt\)

\( \cfrac { dx }{ dt } =\cfrac { 1 }{ m } \psi _{ c }t+\int { \cfrac { u }{ \psi _{ { c } }+u } \cfrac { 1 }{ tanh(x-\cfrac { 1 }{ 2 } x_{ a })-A }  } \cfrac { du }{ dt } du\)

\( \cfrac { d(dx) }{ dudt } =\cfrac { 1 }{ m } \psi _{ c }\cfrac { dt }{ du } +{ \cfrac { u }{ \psi _{ { c } }+u } \cfrac { 1 }{ tanh(x-\cfrac { 1 }{ 2 } x_{ a })-A }  }\cfrac { du }{ dt } \)

\( \cfrac { d(dx) }{ dudt } \cfrac { du }{ dt } =\cfrac { 1 }{ m } \psi _{ c }\cfrac { dt }{ du } \cfrac { du }{ dt } +{ \cfrac { u }{ \psi _{ c }+u } \cfrac { 1 }{ tanh(x-\cfrac { 1 }{ 2 } x_{ a })-A }  }(\cfrac { du }{ dt } )^{ 2 }\)

\( \cfrac { d^{ 2 }x }{ dt^{ 2 } } =\cfrac { 1 }{ m } \psi _{ c }+{ \cfrac { u }{ \psi _{  c  }+u } \cfrac { 1 }{ tanh(x-\cfrac { 1 }{ 2 } x_{ a })-A }  }(\cfrac { du }{ dx } \cfrac { dx }{ dt } )^{ 2 }\)

Finally,

\( \cfrac { d^{ 2 }x }{ dt^{ 2 } } =\cfrac { 1 }{ m } \psi _{ c }+{ \cfrac { u }{ \psi _{ c }+u } { \left( tanh(x-\cfrac { 1 }{ 2 } x_{ a })-A \right)  } }(\cfrac { dx }{ dt } )^{ 2 }\)

where,

\( u=ln(cosh(x-\cfrac { 1 }{ 2 } x_{ a }))-Ax\)

\( \psi _{ c }=\psi _{ n }-\psi _{ max }\)

Compare the above with,

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

Consider,

\( \cfrac { 1 }{ e^{ u }\left( { e^{ 2u }-1 } \right) ^{ 1/2 } } =\cfrac { 1 }{ cosh(x-\cfrac { 1 }{ 2 } x_{ a })sinh(x-\cfrac { 1 }{ 2 } x_{ a }) } ={ 2 }{ csch(2x- x_{ a }) } \)

So,

\( \cfrac { d^{ 2 }x }{ dt^{ 2 } } =\cfrac { 1 }{ m } \psi _{ c }-{ \cfrac { u }{ \psi _{ c }+u }  } { 2 }{ csch(2x- x_{ a }) } (\cfrac { dx }{ dt } )^{ 2 }\)

where \(u=ln(cosh(x-\cfrac { 1 }{ 2 } x_{ a }))\)

for the case where the second integration constant was not considered, (ie \(A=0\)).  Both the terms, in the respective expressions,

\({ 2 }{ csch(2x- x_{ a }) } \)

\(tanh(x-\cfrac { 1 }{ 2 } x_{ a })-A\)

are positive around \(x_a\).

The conclusions drawn previously for \(A=0\) are applicable for \(0\le A\lt1\).

What would cause \(A\ne0\)?  The presence of another particle at a distance \(x\gt2x_z\) or \(x\gt x_a\) where \(x_a\) is the second \(\psi=0\) when \(A=0\).

In the presence of another particle further away, \(\psi\) from the first stretches and envelopes the second particle.  The second particle would behave in a reciprocal way, but in our analysis it was considered a point particle.

Interaction of the particles' \(\psi\) is manifested in the constant \(A\).