It Is Just Nice To Look At

Hei, from the previous post "A Deep Dark Secret" dated 25 Oct 2016,

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

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

$F_{\rho\,n_1}=\cfrac{1}{10\pi^2 }\cfrac{m_{ \psi  }c^2}{a_{\psi\,n1}}$

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,

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

$W_{1\rightarrow 2}=\int^{n2}_{n1}{F_{\rho\,n}}\,d\left(\sqrt[3]{n}.a_{\psi\,c}\right)$

$W_{1\rightarrow 2}=\cfrac{m_{ \psi  }c^2}{10\pi^2 }\int^{n2}_{n1}{\cfrac{1}{\sqrt[3]{n}}}.d\sqrt[3]{n}$

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.