How Does the Optico-Mechanical Analogy Explain Light Behavior in Physics?

[Helios]

So, with the mechanical index of refraction

n = \sqrt{ 1 - V/E }

we plug into the optical ray equation, ( s = arc length )

\nablan - [ \nablan . ( d\vec{r}/ds ) ]( d\vec{r}/ds ) - n ( d^{2} \vec{r}/ ds^{2} ) = 0

and get

\nablaV - [ \nablaV . ( d\vec{r}/ds ) ] ( d\vec{r}/ds ) + 2( E - V )( d^{2} \vec{r}/ ds^{2} ) = 0

Now with the replacements

\vec{F} = -\nablaV

( E - V ) = mv^{2}/2

and get

\vec{F} - [ \vec{F} . ( d\vec{r}/ds ) ] ( d\vec{r}/ds ) - ( mv^{2} )( d^{2} \vec{r}/ ds^{2} ) = 0

d\vec{r}/ds = \hat{T} is a unit vector tangential to the path

d^{2} \vec{r}/ ds^{2} = \hat{N}/R where \hat{N} is a unit normal vector and R is the radius of curvature of the path

So,

\vec{F} - ( \vec{F} . \hat{T} ) \hat{T} - ( mv^{2}/R )\hat{N} = 0

mv^{2}/R is the magnitude of the centripetal force

So with,

\vec{F} = F_{tangent}\hat{T} + F_{normal}\hat{N}

leads me to believe this derivation is correct. Comments?

 

[Helios]

For the derivation of the ray equation see

https://www.physicsforums.com/showthread.php?t=192736

Also here I present the relativistic version of this formulation. I found this and it might be original.