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.
$\blacksquare$
Sources
- 1963: I.M. Gelfand and S.V. Fomin: Calculus of Variations ... (previous) ... (next): $\S 2.9$: The Fixed End Point Problem for n Unknown Functions