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Relativistic mechanical index of refraction

  1. Oct 21, 2007 #1
    I have found the relativistic mechanical index of refraction which I think is

    n = [tex]\sqrt{[( E - V + mc^{2})^{2}- (mc^{2})^{2}]/[( E + mc^{2})^{2}- (mc^{2})^{2}][/tex]

    Follow the same procedure as this thread


    You will have to know that

    mv[tex]^{2}[/tex]/[tex]\sqrt{1 - ( v/c )^{2}}[/tex] = [( E - V + mc[tex]^{2}[/tex] )[tex]^{2}[/tex] - (mc[tex]^{2}[/tex])[tex]^{2}[/tex] ]/( E - V + mc[tex]^{2}[/tex] )

    Also the relativistic centripetal force is

    mv[tex]^{2}[/tex]/R[tex]\sqrt{1 - ( v/c )^{2}}[/tex]

  2. jcsd
  3. Oct 21, 2007 #2


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    I really don't follow the details of this (perhaps other posters will have more to say), but I will point out that dp/dt is also equal to the above expression for a particle of mass m moving at velocity v in a circle of radius R. I.e. the rate of change of momentum of the particle with respect to laboratory time is equal to the gamma*m*v^2/r, with the usual relativistic defintion of gamma.
  4. Oct 23, 2007 #3
    The idea is to turn a kinematic problem into an optical problem. This is the so-called optico-mechanical analogy. The law ( as with the ray theory of light ) says the particle will go from one place to the next in a path of least optical length [tex]\ell[/tex]. This is distinguished from geometric length "s" in which the shortest distance is a straight line. Where there is no potential energy ( V = 0 ), the index of refraction "n" equals 1 and the optical length equals the geometric length and the particle travels in a straight line. Where there is constant potential energy, the particle also travels in a straight line.

    The condition for least optical length is expressed

    [tex]\delta[/tex][tex]\ell[/tex] = [tex]\delta[/tex][tex]\int[/tex] n ds = 0​
    [tex]\delta[/tex][tex]\ell[/tex] = [tex]\int[/tex] [tex]\delta[/tex]n ds + n d[tex]\delta[/tex]s

    [tex]\delta[/tex][tex]\ell[/tex] = [tex]\int[/tex] ( [tex]\nabla[/tex]n . [tex]\delta[/tex][tex]\vec{r}[/tex] ) ds + n ( [tex]\frac{d\vec{r}}{ds}[/tex] . d[tex]\delta[/tex][tex]\vec{r}[/tex] )​

    integrating by parts

    [tex]\delta[/tex][tex]\ell[/tex] = n [tex]\frac{d\vec{r}}{ds}[/tex] . [tex]\delta[/tex][tex]\vec{r}[/tex] + [tex]\int[/tex] ( [tex]\nabla[/tex]n . [tex]\delta[/tex][tex]\vec{r}[/tex] ) ds - d( n [tex]\frac{d\vec{r}}{ds}[/tex] ) . [tex]\delta[/tex][tex]\vec{r}[/tex] ​

    rewrite this as

    [tex]\delta[/tex][tex]\ell[/tex] = n [tex]\delta[/tex]s + [tex]\int[/tex] [ ( [tex]\nabla[/tex]n . [tex]\delta[/tex][tex]\vec{r}[/tex] ) - [tex]\frac{d}{ds}[/tex]( n [tex]\frac{d\vec{r}}{ds}[/tex] ) . [tex]\delta[/tex][tex]\vec{r}[/tex] ]ds ​

    rewrite this as

    [tex]\delta[/tex][tex]\ell[/tex] = n [tex]\delta[/tex]s + [tex]\int[/tex] [ [tex]\nabla[/tex]n - ( [tex]\nabla[/tex]n . [tex]\frac{d\vec{r}}{ds}[/tex])[tex]\frac{d\vec{r}}{ds}[/tex] - n [tex]\frac{d^{2}\vec{r}}{ds^{2}}[/tex] ]. [tex]\delta[/tex][tex]\vec{r}[/tex] ds ​

    There. This is a derivation of the ray equation ( the bracketed expression, which must equal zero ). I see it as a kind of "law of everything", for light and matter. I must comment that I wish it was taught to me in college but it wasn't.

    For the full relevance of this, see the thread

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