\(F=\cfrac { 2 }{ 5 } m_{ \psi }a^{ 2 }_{ \psi }\left( \cfrac { c }{ 2\pi a_{ \psi } } \right) ^{ 2 }\cfrac{1}{a_{\psi}}\)
instead,
\(F_{\rho}=\cfrac { 2 }{ 5 } m_{ \psi }a^{ 2 }_{ \psi }\left( \cfrac { c }{ 2\pi a_{ \psi } } \right) ^{ 2 }\cfrac{1}{a_{\psi}}\)
since we are dealing with \(\psi\) and it is force density all along, \(m_{\psi}=m\) is mass density, a constant. So,
\(F_{\rho\,n_1}=\cfrac { 2 }{ 5 } m_{ \psi }a^{ 2 }_{ \psi \,n1 }\left( \cfrac { c }{ 2\pi a_{ \psi n1 } } \right) ^{ 2 }\cfrac{1}{a_{\psi\,n1}}\)
then
and similarly,
\(F_{\rho\,n_2}=\cfrac{1}{10\pi^2 }\cfrac{m_{ \psi }c^2}{a_{\psi\,n2}}\)
In general,
\(F_{\rho\,n}=\cfrac{1}{10\pi^2 }\cfrac{m_{ \psi }c^2}{a_{\psi\,c}}.\cfrac{1}{\sqrt[3]{n}}\)
since,
\(\cfrac{1}{n}=\left(\cfrac{a_{\psi\,c}}{a_{\psi\,n}}\right)^3\)
as such,
In general,
\(F_{\rho\,n}=\cfrac{1}{10\pi^2 }\cfrac{m_{ \psi }c^2}{a_{\psi\,c}}.\cfrac{1}{\sqrt[3]{n}}\)
since,
\(\cfrac{1}{n}=\left(\cfrac{a_{\psi\,c}}{a_{\psi\,n}}\right)^3\)
as such,
\(W_{1\rightarrow 2}=\int^{a_{ \psi\,n2 }}_{a_{ \psi\,n1 }}{F_{\rho\,n}}\,da_{\psi}\)
\(W_{1\rightarrow 2}=\int^{n2}_{n1}{F_{\rho\,n}}\,d\left(\sqrt[3]{n}.a_{\psi\,c}\right)\)
\(W_{1\rightarrow 2}=\cfrac{m_{ \psi }c^2}{10\pi^2 }\int^{n2}_{n1}{\cfrac{1}{\sqrt[3]{n}}}.d\sqrt[3]{n}\)
and
\(W_{1\rightarrow 2}=\cfrac{m_{ \psi }c^2}{10\pi^2 }\left[ln(\sqrt[3]{n_2})-ln(\sqrt[3]{n_1})\right]=\cfrac{m_{ \psi }c^2}{10\pi^2 }ln\left(\cfrac{\sqrt[3]{n_2}}{\sqrt[3]{n_1}}\right)\)
\(W_{1\rightarrow 2}=\cfrac{m_{ \psi }c^2}{10\pi^2 }ln\left(\cfrac{a_{\psi\,n2}}{a_{\psi\,n1}}\right)\)
Which is somehow very satisfying, irrespective of all and any maths and logic blunders. It is so nice I took a second look. Since,
\(f_c=\cfrac{c}{2\pi a_{\psi\,c}}\)
the expression suggests,
\(h=\cfrac{c}{5\pi }a_{\psi\,c}m_{ \psi }\)
which would be a constant for a given particle. Nice!
A plot of ln((n_2/n_1)^(1/3)) for n_1 = 1 to 7 is given below,
As \(a_{\psi\,n1}\rightarrow a_{\psi\,n2}\), there is an release of energy as the particle grows bigger. \(W_{1\rightarrow 2}\) requires energy and reduces the amount of energy released.
The most energy required from a transition from \(n_1=1\) results in a absorption line.
Big particle with less energy in \(\psi\) is counter-intuitive. Big particle has \(\psi\) going around its circumference at lower frequency, given that \(\psi\) has a constant speed, light speed.
But a big particle contains a small particle!?
A big particle has more extensive \(\psi\) but its \(\psi\) has less energy when we take reference at the circumference/surface of the particle.
\(v=r\omega\)
All inner \(\psi\) have lower speed, when they do not slide along each other. \(\psi\) outside from the center of the particle need to move faster; the fastest of which is light speed.
