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 defintion 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

$$\delta$$$$\ell$$ = $$\int$$ $$\delta$$n ds + n d$$\delta$$s

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

integrating by parts

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

rewrite this as

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

rewrite this as

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