Optico-mechanical analogy

1. Jul 5, 2007

Helios

So, with the mechanical index of refraction

n = $$\sqrt{ 1 - V/E }$$

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

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

and get

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

Now with the replacements

$$\vec{F}$$ = -$$\nabla$$V

( 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$$_{normal}$$$$\hat{N}$$