Geometric Mean And Trickery

Consider a transition from $n_i\rightarrow n_f$,

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

Consider another transition from $a_{ \psi \, f }+a_{ \psi \, i } \rightarrow a_{ \psi \, f }$,  (This is an extrapolation, it is not proven that $a_{ \psi \, f }+a_{ \psi \, i }$ is a solution.)

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

such that, $a_{\psi\,i}$ is in between the paths of $E_{ \Delta h 1}$ and $E_{ \Delta h 2}$.

If we take the geometric mean of the two,

$\bar{E}_{\Delta h}=\sqrt{E_{ \Delta h 1}.E_{ \Delta h 2}}=mc^2\sqrt{\cfrac{a_{ \psi \, i }-a_{ \psi \, f }}{a_{ \psi \, f }+a_{ \psi \, i } }}$

and let

$E_o=-\cfrac{\bar{E}_{\Delta h}}{a_{\psi\,i}}=-mc^2\cfrac{1}{a_{\psi\,i}}\sqrt{\cfrac{a_{ \psi \, i }-a_{ \psi \, f }}{a_{ \psi \, f }+a_{ \psi \, i } }}$  Jm^-1

We have an expression for $E_o$.  A negative sign is necessary because a change from a higher value of $a_\psi$ to a lower value of $a_\psi$ (ie. a positive value of $(a_{ \psi \, i }-a_{ \psi \, f })$ ) results in a loss of energy.

Note:  If,

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

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

And,

$E_{ \Delta h\,i}+E_{ \Delta h\,f}=2\pi a_{\psi\,i}mc.f+2\pi a_{\psi\,f}mc.f$

$2\pi (a_{\psi\,i}+a_{\psi\,f})mc.f=E_{ \Delta h\,i+f}$

And,

$E_{ \Delta h\,i+f}=\cfrac { mc^{ 2 } }{ a_{ \psi \, f }+a_{ \psi \, i } } (a_{ \psi \, f }-(a_{ \psi \, f }+a_{ \psi \, i } ))$ --- (*)

This does not proof that $a_{ \psi \, f }+a_{ \psi \, i }$ is a solution to $x$ for $\psi$ but it shows that (*) is a valid expression.