Main Equation

Proof of Ray Equation. Assume the path is parametrized by $\bf{r}(s) \coloneqq \langle x(s), y(s), z(s) \rangle$. Using

Theorem (Euler-Lagrange Equation).

For a functional of the form

$$\int L(x, y, z, x^{'}, y^{'}, z^{'}) \ \mathrm{d}s,$$

the path that minimizes/maximizes the integral is determined by the following equations:

$$\frac{d}{ds}\left(\frac{\partial L}{\partial x'}\right) - \frac{\partial L}{\partial x} = 0 \quad \text{(and similarly for } y, z\text{)}.$$

on Fermat’s Principle, we account for the arc-length constraint $\left(\frac{dx}{ds}\right)^2 + \left(\frac{dy}{ds}\right)^2 + \left(\frac{dz}{ds}\right)^2 = 1$ via Lagrange multipliers. The augmented Lagrangian is:

$$\tilde{L} = n(\mathbf{r}) + \lambda(s) \left[\left(\frac{dx}{ds}\right)^2 + \left(\frac{dy}{ds}\right)^2 + \left(\frac{dz}{ds}\right)^2 - 1\right]$$

Computing $\frac{\partial \tilde{L}}{\partial x’} = 2\lambda(s) \frac{dx}{ds}$ and applying the Euler-Lagrange equation, we get:

$$\frac{d}{ds}\left(2\lambda(s) \frac{dx}{ds}\right) = \frac{\partial n}{\partial x}$$

For the actual ray trajectory, the Lagrange multiplier identifies with the refractive index: $\lambda(s) = \frac{n(\mathbf{r}(s))}{2}$. This yields three separate PDEs:

$$\begin{align} \frac{\mathrm{d}}{\mathrm{d}s} \left( n \frac{\mathrm{d}x}{\mathrm{d}s} \right) &= \frac{\partial n}{\partial x} \\ \frac{\mathrm{d}}{\mathrm{d}s} \left( n \frac{\mathrm{d}y}{\mathrm{d}s} \right) &= \frac{\partial n}{\partial y} \\ \frac{\mathrm{d}}{\mathrm{d}s} \left( n \frac{\mathrm{d}z}{\mathrm{d}s} \right) &= \frac{\partial n}{\partial z} \end{align}$$

These three equations can be compactly written as the vector equation: $\frac{d}{ds}\left(n\frac{d\mathbf{r}}{ds}\right) = \nabla n$.

• In general, the way you would solve the Ray Equation is by describing the trajectory by two functions and writing $\mathrm{d}s = \mathrm{d}z \sqrt{ 1 + \left( \frac{\mathrm{d}x}{\mathrm{d}z} \right)^2 + \left( \frac{\mathrm{d}y}{\mathrm{d}z} \right)^2 }$

Paraxial Regime