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Optico-mechanical analogy

  1. Jul 5, 2007 #1
    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?
     
  2. jcsd
  3. Oct 26, 2007 #2
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