# Propagation of Light in Inhomogeneous Medium

## Theorem

Let $v: \R^3 \to \R$ be a real function.

Let $M$ be a 3-dimensional Euclidean space.

Let $\gamma:t \in \R \to M$ be a smooth curve embedded in $M$, where $t$ is time.

Denote its derivative with respect to time by $v$.

Suppose $M$ is filled with an optically inhomogeneous medium such that at each point speed of light is $v = \map v {x, y, z}$

Suppose $\map y x$ and $\map z x$ are real functions.

Let the light move according to Fermat's principle.

Then equations of motion have the following form:

$\dfrac {\partial v} {\partial y} \dfrac {\sqrt {1 + y'^2 + z'^2} } {v^2} + \dfrac \d {\d x} \dfrac {y'} {v \sqrt {1 + y'^2 + z'^2} } = 0$
$\dfrac {\partial v} {\partial z} \dfrac {\sqrt {1 + y'^2 + z'^2} } {v^2} + \dfrac \d {\d x} \dfrac {z'} {v \sqrt {1 + y'^2 + z'^2} } = 0$

## Proof

By assumption, $\map y x$ and $\map z x$ are real functions.

This allows us to $x$ instead of $t$ to parameterize the curve.

This reduces the number of equations of motion to $2$, that is: $\map y x$ and $\map z x$.

The time it takes to traverse the curve $\gamma$ equals:

 $\ds T$ $=$ $\ds \int_{t_a}^{t_b} \rd t$ $\ds$ $=$ $\ds \int_{s_a}^{s_b} \dfrac {\d t} {\d s} \rd s$ Chain Rule for Derivatives, $\d s$ - arc length element $\ds$ $=$ $\ds \int_a^b \dfrac 1 v \frac {\d s} {\d x} \rd x$ Chain Rule for Derivatives, $v = \dfrac {\d s} {\d t}$ $\ds$ $=$ $\ds \int_a^b \frac {\sqrt{1 + y'^2 + z'^2} } {\map v {x, y, z} } \rd x$ Definition of Arc Length in 3-dimensional Euclidean space

According to Fermat's principle, light travels along the trajectory of least time.

Therefore, this integral has to be minimized with respect to $\map y x$ and $\map z x$.

It holds that:

$\dfrac \partial {\partial y} \dfrac {\sqrt {1 + x'^2 + y'^2} } {\map v {x, y, z} } = -\dfrac {\sqrt {1 + x'^2 + y'^2} } {\map {v^2} {x, y, z} } \dfrac {\partial v} {\partial y}$

Also:

$\dfrac \d {\d x} \dfrac \partial {\partial y'} \dfrac {\sqrt{1 + x'^2 + y'^2} } {\map v {x, y, z} } = \dfrac \d {\d x} \dfrac {y'} {v \sqrt {1 + x'^2 + y'^2} }$

Analogous relations hold for $\map z x$.

Then, by Euler's Equations and multiplication by $-1$, the desired result follows.

$\blacksquare$