Planck's Constant Dead

In the case when $n=1$,  as $\psi$ goes around the radius of the celestial body, $r=a_{\psi\,f}$, one photon pulse is emitted and between two photons, an EMP.

The frequency is simply,

$f=\cfrac{c}{2\pi r}=\cfrac{c}{2\pi a_{\psi\,f}}$

The question of whether photon/EMP emission occurs at at peak $A_f$, when a transition to $A_i=0$ results in a collapse of the amplitude, or when at $A_i$, $\psi$ returns all a quarter of the period to $A_f=0$, is still open.  Either scenario is possible.  Together with the all possible conditions for emission as amplitude, $A_f$ increases (Post "Pop Now, Pop Later" dated 06 Oct 2017) invites other possible emission frequencies,

$f=\cfrac{c}{2\pi r}=\cfrac{c}{2\pi a_{\psi\,i}}$ 

when $r=a_{\psi_i}$ is the emission condition.  These emitted frequencies do not indicate the difference in energy levels that, on transition between them, leads to the emission. Only when (if) the emission occurs on the criterion of equal perimeter,

$f=\cfrac{c}{2\pi r}=\cfrac{c}{2\pi \sqrt{2a^2_{\psi\,i}-a^2_{\psi\,f}}}$

does the expression for frequency/energy of the quanta, photon or EMP, reflect the different between the energy levels as shown by the term,

$\cfrac{1}{\sqrt{2a^2_{\psi\,i}-a^2_{\psi\,f}}}$

Notice the factor $2$ before $a^2_{\psi\,i}$.

The difference in energy levels, when a emission occurs, also shows up in $A_n$. 

$A_f=\sqrt{2}\sqrt{a^2_{\psi\,i}-a^2_{\psi\,f}}$

In the case of  of an emission when $r=a_{\psi\,i}$ where the radius of the elliptical path touches $a_{\psi\,i}$,

$A_f=\sqrt{a^2_{\psi\,i}-a^2_{\psi\,f}}$

These expressions differ by a constant factor of $\sqrt{2}$.  Since,

$\cfrac{1}{n}=\left(\cfrac{a_{\psi\,c}}{a_{\psi\,n}}\right)^3$

$a_{\psi\,n}=\sqrt[3]{n}.a_{\psi\,c}$

and from the post "Touch And Go" dated 24 Dec 2014,

$E_{ o }=-mc^{ 2 }\cfrac { 1 }{ a_{ \psi \, i } } \sqrt { \cfrac { (a_{ \psi \, i}-a_{ \psi \, f }) }{ (a_{ \psi \, i }+a_{ \psi \, f }) }  }$

and,

$E_{ \Delta n }=E_{ o }\sqrt { a^{ 2 }_{ \psi \, i}-a^{ 2 }_{ \psi \, f} }$

$E_{ \Delta n }=mc^{ 2 }(\cfrac { a_{ \psi \, f } }{ a_{ \psi \, i } } -1)$

$E_{ \Delta n }=mc^{ 2 }(\sqrt[3]{\cfrac{n_f}{n_i}} -1)=mc^{ 2 }(\cfrac{\sqrt[3]{n_f}-\sqrt[3]{n_i}}{\sqrt[3]{n_i}})$

on a transition from $n_f\rightarrow n_i$.  where $n_f$ is larger then $n_i$ when $a_{\psi\,f}$ is smaller than $a_{\psi\,i}$.  What is this expression about?  Consider,

$f\lambda=c=f_c.2\pi a_{\psi\,c}=f_c.2\pi\cfrac{a_{\psi\,f}}{\sqrt[3]{n_f}}$

we have,

$E_{ \Delta n }=mc.f_c.2\pi a_{\psi\,f}\left(\cfrac{\sqrt[3]{n_f}-\sqrt[3]{n_i}}{\sqrt[3]{n_i}\sqrt[3]{n_f}}\right)$

where $mc.f_c$ is the rate of change of momentum over one period.  It is a force along the orbital path of $\psi$ and $2\pi a_{\psi\,f}$ is the total distance along such this path.  That is to say,

$mc.f_c.2\pi a_{\psi\,f}=work\,\,done$

is the work done along the orbital path as $\psi$ moves at light speed, $c$ around an orbit of radius $a_{\psi\,f}$.

If we set,

$E=h.f_c\left(\cfrac{\sqrt[3]{n_f}-\sqrt[3]{n_i}}{\sqrt[3]{n_i}\sqrt[3]{n_f}}\right)$

then,

$h=mc.2\pi a_{\psi\,f}$

on a transition from $n_f\rightarrow n_i$.  Given $n_f$, $h$ is a constant.

We have this result before.  Maybe Planck constant is also dead...  罗刹 of all constants.