Path of light in material of continuously varying refractive index

[etotheipi]

Homework Statement

Suppose that the speed of light $c(y)$ varies continuously through a medium and is a function of the distance from the boundary $y=0$. Use Fermat's principle to show that the path $y(x)$ of the light ray is given by

$$c(y)y'' + c'(y)(1+y'^{2})=0$$

Relevant Equations

N/A

I've been playing around with this for quite some time now this morning but can't get the last bit out. I defined the time functional to be $$T[y] = \int_{x_1}^{x_2} \frac{\sqrt{1+(y')^{2}}}{c(y)} dx$$ which follows from consideration of the time taken to cover an infinitesimal section of arc. I want to find the $y(x)$ that minimises $T$ so I let $F(x,y,y') = \frac{\sqrt{1+(y')^{2}}}{c(y)}$ and put the whole thing into the magic equation $$
\begin{align}
\frac{d}{dx}\left(\frac{\partial F}{\partial y'}\right) &= \frac{\partial F}{\partial y} \\ \frac{d}{dx} \left( \frac{1}{c(y)} \frac{y'}{\sqrt{1+(y')^{2}}} \right) &= \frac{\partial F}{\partial y} \\ \frac{1}{c(y)} \frac{y''\sqrt{1+(y')^{2}} - \frac{(y')^{2}}{\sqrt{1+(y')^2}}}{1+(y')^2} + \frac{c'(y)}{c(y)^{2}} \frac{(y')^{2}}{\sqrt{1+(y')^2}} &= - \sqrt{1+ (y')^2} \frac{c'(y)}{c(y)^{2}}
\end{align}
$$ This turns out to be equivalent to $$c(y) y'' + c'(y) (1+(y')^2) = \frac{c(y)(y')^2}{1+(y')^2} - c'(y)(1+(y')^2)$$So it's sort of what we want, except not really since I've got that annoying thing on the RHS! Any help would be appreciated!

 

[etotheipi]

I figured it out, there were two errors in my working!

In equation (3), I ended the derivative chain too early and forgot to multiply the second term in the first fraction by $y''$. The first fraction should actually read $$\frac{1}{c(y)} \frac{y''\sqrt{1+(y')^{2}} - \frac{(y')^{2}(y'')}{\sqrt{1+(y')^2}}}{1+(y')^2}$$ And secondly, the second fraction in equation (3) should be negated. With those corrections, the desired result is obtained after a bit of cancellation.

Phew...

 

[Fred Wright]

With all due respect, I have a different take on your problem. Your optical Lagrangian is, $$
\mathcal L=\frac{\sqrt{(1+y'^2)}}{c(y)}$$ and thus$$
\frac{\partial{\mathcal L}}{\partial {y'}}=\frac{y'}{c(y)\sqrt{(1+y'^2)}}\\
\frac{d}{dx}(\frac{\partial{\mathcal L}}{\partial {y'}})=\frac{y''}{c(y)\sqrt{(1+y'^2)}} + y'\frac{d}{dx}(\frac{1}{c(y)\sqrt{(1+y'^2)}})\\
\frac{\partial{\mathcal L}}{\partial {y}}=-\frac{c'(y)\sqrt{(1+y'^2)}}{c(y)^2}$$ and we have.$$
\frac{y''}{c(y)\sqrt{(1+y'^2)}}+y'\frac{d}{dx}(\frac{1}{c(y)\sqrt{(1+y'^2)}})=-\frac{c'(y)\sqrt{(1+y'^2)}}{c(y)^2}\\
c(y)y''+c'(y)(1+y'^2)=-c(y)y'\sqrt{(1+y'^2)}\frac{d}{dx}(\frac{1}{c(y)\sqrt{(1+y'^2)}})$$For the l.h.s. to be zero, we must have,$$
\frac{d}{dx}(\frac{c}{c(y)\sqrt{(1+y'^2)}})=0$$where I have multiplied by c (speed of light in vacuum) and we see $\frac{c}{c(y)\sqrt{(1+y'^2)}}$ is a constant of motion. The ratio $\frac{c}{c(y)}$ is the index of refraction and we infer that the index of refraction is proportional to a differential element of arc length.

 

[etotheipi]

For the l.h.s. to be zero, we must have,$$
\frac{d}{dx}(\frac{c}{c(y)\sqrt{(1+y'^2)}})=0$$where I have multiplied by c (speed of light in vacuum) and we see $\frac{c}{c(y)\sqrt{(1+y'^2)}}$ is a constant of motion. The ratio $\frac{c}{c(y)}$ is the index of refraction and we infer that the index of refraction is proportional to a differential element of arc length.

That's a really nice insight, I hadn't considered anything further than the given result!