- #1
Helios
- 269
- 63
So, with the mechanical index of refraction
n = [tex]\sqrt{ 1 - V/E }[/tex]
we plug into the optical ray equation, ( s = arc length )
[tex]\nabla[/tex]n - [ [tex]\nabla[/tex]n . ( d[tex]\vec{r}[/tex]/ds ) ]( d[tex]\vec{r}[/tex]/ds ) - n ( d[tex]^{2}[/tex] [tex]\vec{r}[/tex]/ ds[tex]^{2}[/tex] ) = 0
and get
[tex]\nabla[/tex]V - [ [tex]\nabla[/tex]V . ( d[tex]\vec{r}[/tex]/ds ) ] ( d[tex]\vec{r}[/tex]/ds ) + 2( E - V )( d[tex]^{2}[/tex] [tex]\vec{r}[/tex]/ ds[tex]^{2}[/tex] ) = 0
Now with the replacements
[tex]\vec{F}[/tex] = -[tex]\nabla[/tex]V
( E - V ) = mv[tex]^{2}[/tex]/2
and get
[tex]\vec{F}[/tex] - [ [tex]\vec{F}[/tex] . ( d[tex]\vec{r}[/tex]/ds ) ] ( d[tex]\vec{r}[/tex]/ds ) - ( mv[tex]^{2}[/tex] )( d[tex]^{2}[/tex] [tex]\vec{r}[/tex]/ ds[tex]^{2}[/tex] ) = 0
d[tex]\vec{r}[/tex]/ds = [tex]\hat{T}[/tex] is a unit vector tangential to the path
d[tex]^{2}[/tex] [tex]\vec{r}[/tex]/ ds[tex]^{2}[/tex] = [tex]\hat{N}[/tex]/R where [tex]\hat{N}[/tex] is a unit normal vector and R is the radius of curvature of the path
So,
[tex]\vec{F}[/tex] - ( [tex]\vec{F}[/tex] . [tex]\hat{T}[/tex] ) [tex]\hat{T}[/tex] - ( mv[tex]^{2}[/tex]/R )[tex]\hat{N}[/tex] = 0
mv[tex]^{2}[/tex]/R is the magnitude of the centripetal force
So with,
[tex]\vec{F}[/tex] = F[tex]_{tangent}[/tex][tex]\hat{T}[/tex] + F[tex]_{normal}[/tex][tex]\hat{N}[/tex]
leads me to believe this derivation is correct. Comments?
n = [tex]\sqrt{ 1 - V/E }[/tex]
we plug into the optical ray equation, ( s = arc length )
[tex]\nabla[/tex]n - [ [tex]\nabla[/tex]n . ( d[tex]\vec{r}[/tex]/ds ) ]( d[tex]\vec{r}[/tex]/ds ) - n ( d[tex]^{2}[/tex] [tex]\vec{r}[/tex]/ ds[tex]^{2}[/tex] ) = 0
and get
[tex]\nabla[/tex]V - [ [tex]\nabla[/tex]V . ( d[tex]\vec{r}[/tex]/ds ) ] ( d[tex]\vec{r}[/tex]/ds ) + 2( E - V )( d[tex]^{2}[/tex] [tex]\vec{r}[/tex]/ ds[tex]^{2}[/tex] ) = 0
Now with the replacements
[tex]\vec{F}[/tex] = -[tex]\nabla[/tex]V
( E - V ) = mv[tex]^{2}[/tex]/2
and get
[tex]\vec{F}[/tex] - [ [tex]\vec{F}[/tex] . ( d[tex]\vec{r}[/tex]/ds ) ] ( d[tex]\vec{r}[/tex]/ds ) - ( mv[tex]^{2}[/tex] )( d[tex]^{2}[/tex] [tex]\vec{r}[/tex]/ ds[tex]^{2}[/tex] ) = 0
d[tex]\vec{r}[/tex]/ds = [tex]\hat{T}[/tex] is a unit vector tangential to the path
d[tex]^{2}[/tex] [tex]\vec{r}[/tex]/ ds[tex]^{2}[/tex] = [tex]\hat{N}[/tex]/R where [tex]\hat{N}[/tex] is a unit normal vector and R is the radius of curvature of the path
So,
[tex]\vec{F}[/tex] - ( [tex]\vec{F}[/tex] . [tex]\hat{T}[/tex] ) [tex]\hat{T}[/tex] - ( mv[tex]^{2}[/tex]/R )[tex]\hat{N}[/tex] = 0
mv[tex]^{2}[/tex]/R is the magnitude of the centripetal force
So with,
[tex]\vec{F}[/tex] = F[tex]_{tangent}[/tex][tex]\hat{T}[/tex] + F[tex]_{normal}[/tex][tex]\hat{N}[/tex]
leads me to believe this derivation is correct. Comments?