\(L=I\omega \)
\( F=L\cfrac{\omega}{r} \)
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,
\( \psi =-{ 2{ mc^{ 2 } } }\, ln(cosh(\cfrac { G }{ \sqrt { 2{ mc^{ 2 } } } } (a_{ \psi })))\)
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!
