Success Is In The Way You Handle The Question

There is a problem, if

$\psi_r=h.f_r=\psi_o+x_rF=2\pi a_{\psi\,r}mc$

then for

$\psi_o\le\psi_r$

$a_{\psi\,r}\ge a_{\psi\,o}$

so,

$a_{\psi\,r}\ne a_{\psi\,o}-d$

but,

$a_{\psi\,o}= a_{\psi\,r}-d$

????

But more refracted rays has higher frequency.

In the case of,

$h.f=2\pi a_\psi.mc$

as $a_\psi\,\downarrow$, since

$2\pi a_\psi=\lambda$

$a_\psi\,\downarrow\implies\lambda\,\downarrow$

but,

$f\lambda=c$

so, $f\,\uparrow$

This is consistent with smaller $a_\psi$ having higher frequency, $f$.

This could mean that,

$h.f=2\pi a_\psi.m_fc$

that, $m_f$ the mass of the photon changes with changing $a_{\psi}$, $\psi=h.f$ being constant as $a_\psi$ deform.  This is consistent with the idea that mass, $m$ of a particle and its surrounding $\psi$ are interchangeable as discussed in the post "We Still Have A Problem" and "Less Mass But No Theoretical Mass" dated 23 Nov 2014.  For this case, a reduction in $\psi$ leads to an decrease in mass, $m$.  As such, $m_{a_{\psi\,o}}\ne m_{a_{\psi\,r}}$ along with a change in $a_{\psi}$.

and (!!!)

$d\lt 0$

that $\psi$ tends to disperse in a denser medium.  That a decrease in energy density $\psi$,

$\psi_{n\,r}\lt\psi_{n\,i}$

is an increase in geometric size of the $\psi$ sphere,

$d\lt0$  as  $a_{\psi\,o}\lt a_{\psi\,r}$

that the $\psi$ sphere has negative deformation/compression on impact.  How did $\psi_n$ decrease in the first place, on impact?  This was deduced from Snell's Law in the post "Photons Like Bubbles" dated 14 Jun 15.

With these in mind, how much of the previous discussion is still valid?

$n=\cfrac{a_\psi}{a_\psi-d}$ is wrong!

from,

$n=\cfrac{\psi_{n\,i}}{\psi_{n\,r}}$

A more appropriate expression is,

$n=\cfrac{a_{\psi\,o}}{a_{\psi\,o}-d}.\cfrac{m_{a_{\psi\,o}}}{m_{a_{\psi\,r}}}$

from,

$n=\cfrac{\psi_{n\,i}}{\psi_{n\,r}}=\cfrac{2\pi a_{\psi\,o}m_{a_{\psi\,o}}}{2\pi a_{\psi\,r}m_{a_{\psi\,r}}}$

and

$a_{\psi\,r}=a_{\psi\,o}-d$

The masses, $m_{a_{\psi\,o}}$ and $m_{a_{\psi\,r}}$ before and after refraction are different due to the reduction in $\psi_n$.

Since,

$h.f_o=h.\cfrac{c}{\lambda_o}=h.\cfrac{c}{2\pi a_{\psi\,o}}=2\pi a_{\psi\,o}m_{a_{\psi\,o}}$

we have,

$1=\cfrac{a^2_{\psi\,o}m_{a_{\psi\,o}}}{a^2_{\psi\,r}m_{a_{\psi\,r}}}$

as  $\cfrac{hc}{(2\pi)^2}$ is a constant.

And the expression for $n$ becomes,

$n=\cfrac{a_{\psi\,o}}{a_{\psi\,o}-d}.\cfrac{m_{a_{\psi\,o}}}{m_{a_{\psi\,r}}}=\cfrac{a_{\psi\,o}}{a_{\psi\,r}}.\cfrac{m_{a_{\psi\,o}}}{m_{a_{\psi\,r}}}$

as

$a_{\psi\,o}-d=a_{\psi\,r}$

becomes,

$n=\cfrac{a_{\psi\,r}}{a_{\psi\,o}}=\cfrac{a_{\psi\,o}-d}{a_{\psi\,o}}$

remember that $d\lt0$.  So,

$d=a_{\psi\,o}(1-n)$

From the post "Success Is In The Way You Handle The Unknown",

$A=a_{\psi\,o}^{1/2}d^{3/2}$

$A$ is also negative as $d$ is negative.

Substitute in $d$,

$A=a^2_{\psi\,o}(1-n)^{3/2}$

$n=1-\cfrac { A^{ 2/3 } }{ a^{ 4/3 }_{\psi\,o} }$

$n=1+\cfrac { B }{ a^{ 4/3 }_{\psi\,o} }$

where $B=-A^{2/3}$ is positive.  As,

$\lambda_o=2\pi a_{\psi\,o}$

we may have,

$n=1+\cfrac { C }{ \lambda^{ 4/3 }_{o} }$

where $C=(2\pi)^{4/3}B$ is a positive constant.  Unfortunately, still not Sellmeier equation.

Furthermore we have, because of the introduction of $m_{a_\psi}$,

$h.f_d=2\pi a_\psi.m_{a_\psi}c+\cfrac{8}{15}E^*a_\psi^{1/2}d^{5/2}$

we can still obtain $E^*$ from the set of gradients obtained by varying $a_\psi$ and a relationship in the change of photon mass, $m_{a_\psi}$ with $a_\psi$ from the y-intercepts.

There is still a reduction in $\psi_n$ in the direction of impact as $\psi$ enters into the medium, but such a decrease in $\psi$ corresponds physically to an increase in geometric dimension ($-d$ is positive).  $\psi$ burgle at the point of impact.  Intuitively, this make sense because $\psi$ is after all a density term.  A local increase in size reduces density.

Is it double counting to introduce both $m_{a_\psi}$ and $-d$ ?  No.