A Deep Dark Secret

Consider the force holding $\psi$ in circular motion,

$L=I\omega$

$F=L\cfrac{\omega}{r}$

scale by $r$, as the further from the center the less $L$ has to turn,

$F=I\cfrac{\omega ^{ 2 }}{r}$

In the case of a point mass, $m$ in circular motion with velocity $v$, in circle of radius of $r$,

$F=mr^2\left(\cfrac{v}{r}\right)^2.\cfrac{1}{r}=m\cfrac{v^2}{r}$

where $I=mr^2$ and  $\omega =\cfrac{v}{r}$ in radian per second.  Which is as expected.

In this case of a sphere of $\psi$,

$F=\cfrac { 2 }{ 5 } m_{ \psi  }a^{ 2 }_{ \psi  }\left( \cfrac { c }{ 2\pi a_{ \psi  } }  \right) ^{ 2 }.\cfrac{1}{a_{ \psi  } }$

where $\omega=\cfrac{v}{2\pi a_{\psi}}$ in per second.

$m_{ \psi  }=\rho _{ \psi  }.\cfrac { 4 }{ 3 } \pi a^{ 3 }_{ \psi  }$

$F=\cfrac { 4 }{ 30\pi }  c^{ 2 }\rho _{ \psi  }.a^{ 2 }_{ \psi  }$

$W_{ 1\rightarrow 2 }=\int _{ a_{ \psi \,n1 } }^{ a_{ \psi \, n2 } }{ F } da_{ \psi  }$

If we assume that, $\rho _{ \psi  }\propto \psi$.

For,

$a_{ \psi \, n1 }>a_{ \psi \,\pi  }$ and $a_{ \psi \, n2 }>a_{ \psi \,\pi  }$

$\psi=\psi_{\pi}=constant$

We have,

$W_{ 1\rightarrow 2 }=\cfrac { 4 }{ 90\pi } c^{ 2 }\rho _{ \psi  }\left[ a^{ 3 }_{ \psi \, n_{ 2 } }-a^{ 3 }_{ \psi \, n_{ 1 } } \right]$

$W_{ 1\rightarrow 2 }=\cfrac { 4 }{ 90\pi  } c^{ 2 }\rho _{ \psi  }a^{ 3 }_{ \psi \, c }\left[ n_{ 2 }-n_{ 1 } \right]$

which is just as wrong as the other expressions derived previously.

For,

$a_{ \psi \, \, n1 }\lt a_{ \psi \, \, \pi  }$

$\, \, a_{ \psi \, \, n2 }\lt a_{ \psi \, \, \pi  }$

$W_{ 1\rightarrow 2 }=\int _{ a_{ \psi \, \, n1 } }^{ a_{ \psi \, \, n2 } }{ F } da_{ \psi  }=\cfrac { 4 }{ 30 \pi} c^{ 2 }\int _{ a_{ \psi \, \, n1 } }^{ a_{ \psi \, \, n2 } }{ \rho _{ \psi  }.a^{2 }_{ \psi  } } da_{ \psi  }$

From the post "Not Quite The Same Newtonian Field" dated 23 Nov 2014,

$\psi =-i{ 2{ mc^{ 2 } } }\, ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } (x-x_{ z })))+c$

$\psi =-{ 2{ mc^{ 2 } } }\, ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } (a_{ \psi  })))$

under the assumption, $\rho _{ \psi  }\propto \psi$

$\rho _{ \psi  }=A. \psi$

We have,

$W_{ 1\rightarrow 2 }=A.\cfrac { 8 }{ 30 \pi } mc^{ 2 }\int _{ a_{ \psi \,  n1 } }^{ a_{ \psi \,  n2 } }{ -a^{ 2 }_{ \psi  } } ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } (a_{ \psi  })))\,da_{ \psi  }$

At last, inevitably the ugly bride meets the in-laws ...

What is $m$?  This was a problem since $F_{\rho}$ or $F$, the force density was written down.  A force has to act on some mass.  If force density acts on mass density then,

$\rho_{\psi}=m$

then $\psi=f(\rho_{\psi})$, given $\rho_\psi$,

$\psi =-{ 2{ \rho_{\psi}c^{ 2 } } }\, ln(cosh(\cfrac { G }{ \sqrt { 2{ \rho_{\psi}c^{ 2 } } }  } (a_{ \psi  })))$

from which we solve for $\rho_{\psi}$ given $\psi$.  Which make sense because $W_{ 1\rightarrow 2 }$ is moving $\psi$ about, we must know the amount of $\psi$ in question to calculate $W_{ 1\rightarrow 2 }$.

But does energy density has mass density?  Does energy has mass?

This path does not provide an easy answer to $W_{ 1\rightarrow 2 }$  as $L_{n1}\rightarrow L_{n2}$.  $W_{ 1\rightarrow 2 }$ might be similar to Rydberg formula,

$\cfrac{1}{\lambda}=R_{\small{H}}\left(\cfrac{1}{n^2_1}-\cfrac{1}{n^2_2}\right)$

A new passport and a ticket to nowhere...you asked for it!