Relativistic mechanical index of refraction

[Helios]

I have found the relativistic mechanical index of refraction which I think is

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

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

Also the relativistic centripetal force is


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

.

 

[pervect]

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 definition of gamma.

 

[Helios]

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 \ell. 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

\delta\ell = \delta\int n ds = 0​

then

\delta\ell = \int \deltan ds + n d\deltas

\delta\ell = \int ( \nablan . \delta\vec{r} ) ds + n ( \frac{d\vec{r}}{ds} . d\delta\vec{r} )​

integrating by parts

\delta\ell = n \frac{d\vec{r}}{ds} . \delta\vec{r} + \int ( \nablan . \delta\vec{r} ) ds - d( n \frac{d\vec{r}}{ds} ) . \delta\vec{r}​

rewrite this as

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

rewrite this as

\delta\ell = n \deltas + \int [ \nablan - ( \nablan . \frac{d\vec{r}}{ds})\frac{d\vec{r}}{ds} - n \frac{d^{2}\vec{r}}{ds^{2}} ]. \delta\vec{r} 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