Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    https://www.physicsforums.com/showthread.php?t=176081&highlight=optico-mechanical

    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

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    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​
    then
    [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

    https://www.physicsforums.com/showthread.php?t=176081&highlight=optico-mechanical
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Relativistic mechanical index of refraction
  1. Relativistic Mechanics (Replies: 5)

  2. Relativistic Mechanics (Replies: 19)

Loading...