From wave optics to geometrical optics

In this post, we will show how the wave optics becomes the geometrical optics in the limit of zero wavelength. When in the university, I only took one optics course at the level of college physics, which only mentioned the concept of Fermat's principle without too much discussions.

Fermat's principle

In geometrical optics, the Fermat's principle states that the path $\vec{r}$ of a light ray from the point A to B always minimizes the quantity: \begin{equation}\mathbb{S}[\vec{r}] = \int_A^B n(\vec{r})\,ds\,, \tag{1}\end{equation} where $n(\vec{r})$ is the distribution of reflective index of the medium in space and $ds$ is the differential arc length along the path \begin{equation}ds =\sqrt{d\vec{r}\cdot d\vec{r}}\,.\end{equation} The Fermat's principle provides an explicit connection between the geometrical optics and the classical mechanics in the form of principle of least action.

Warm up: Snell's law

Before deriving the general equation of motion from the Fermat's principle, we do a simple exercise as the warm up. Consider a two-dimensional space, the differential arc length in in Cartesian coordinates $(x, y)$ is $ds=\sqrt{\left(dx\right)^2 + \left(dy\right)^2}$ and the Fermat's principle can be written as \begin{equation}\mathbb{S}[y(x)] = \int_A^B n(y) \sqrt{\left(dx\right)^2 + \left(dy\right)^2}=\int_A^B n(x, y) \sqrt{1 + {y'(x)}^2}dx\,,\end{equation} which is a functional of the path described by the function $y(x)$. Compared to the action in the classical mechanics, as reviewed in this post, the Lagrangian in this case is \begin{equation} L(y, y', x)=n(x, y)\sqrt{1+ {y'(x)}^2}\,,\end{equation} and the Euler-Lagrange equation is $\frac{d}{dx}\frac{\partial L}{\partial y'}-\frac{\partial L}{\partial y}=0$, i.e. \begin{equation}\frac{d}{dx}\left(n(x, y)\frac{y'(x)}{\sqrt{1+ {y'(x)}^2}}\right)- \sqrt{1+ {y'(x)}^2}\frac{\partial n(x, y)}{\partial y}=0\,.\end{equation} Now suppose the reflective index is uniform along $y$-axis, i.e. $\partial_y n =0$, the above equation becomes \begin{equation} n( x)\frac{y'(x)}{\sqrt{1+ {y'(x)}^2}} = \text{const.} \end{equation} Let $y'(x)\equiv \tan \theta_x$, where $\theta$ is the local angle between the path and the $x$-axis (the normal of boundary between different reflective index), we have \begin{equation}n(x) \sin \theta_x=\text{const.}\end{equation} So we discovered the Snell's law from the Fermat's principle!

General equation of motion

In general, we can parametrize a path by some parameter $\sigma$, then the Fermat's principle (1) is of the form \begin{equation}\mathbb{S}[\vec{r}(\sigma)] = \int_A^B n(\vec{r}(\sigma))\,\sqrt{\frac{d\vec{r}(\sigma)}{d\sigma}\cdot\frac{d\vec{r}(\sigma)}{d\sigma}}d\sigma\,.\end{equation} The Lagrangian in this case is $L(\vec{r}, {\vec{r}}', \sigma)=n(\vec{r})\sqrt{{\vec{r}}'\,\cdot\, {\vec{r}}'}$ and the Euler-Lagrangian equation is \begin{equation}\frac{d}{d\sigma}\left[\frac{n(\vec{r})}{\sqrt{{\vec{r}}'\cdot {\vec{r}}'}}\frac{d\vec{r}}{d\sigma}\right] -\sqrt{{\vec{r}}'\cdot {\vec{r}}'}\nabla n(\vec{r})=0\,.\end{equation} Among all choices of the parameter $\sigma$, there is a natural choice: the arc length of the path $s$ because of a nice property $\left|\frac{d\vec{r}}{ds}\right|=1$. The above equation can then be simplified as \begin{equation}\boxed{\frac{d}{ds}\left[n(\vec{r})\frac{d\vec{r}}{ds}\right] -\nabla n(\vec{r})=0}\,,\tag{2}\end{equation} which is a general equation of motion of a light ray.

BTW, Newton believes that light is a group of particles. Thanks for Hamiltonian! the Fermat's principle in geometrical optics and the principle of least action in classical mechanics are the same, at least mathematically. The light ray's equation (2) can also be written in the form of Newton's second law \begin{equation}\frac{d^2 r}{d\tau^2} = \nabla \left(\frac{n(\vec{r})^2}{2}\right)\,\end{equation} by letting $ds=n\,d\tau$ in Eq. (2). The "potential" driving the motion of "particles" of the light ray is $\frac{1}{2}n^2$.

From waves to rays

Classically, lights are actually electromagnetic waves described by the Maxwell's equation. Here for simplicity, we consider a scalar $\psi$ rather than the electric and magnetic fields. The wave equation is \begin{equation} \nabla ^2 \psi(\vec{r},t)- \frac{n(\vec{r})^2}{c^2}\frac{\partial ^2 }{\partial t^2}\psi(\vec{r},t)=0\,.\tag{3}\end{equation} We now prove that the wave equation (3) reduces to the ray equation (2) when the wavelength goes to zero. For this purpose, we start by the ansatz \begin{equation}\psi(\vec{r},t) = \rho(\vec{r}) e^{ik(S(\vec{r})-ct)}\,,\end{equation} where $k\equiv\frac{2\pi}{\lambda}$. We plug the ansatz into the wave equation (3) by computing \begin{eqnarray}\nabla \psi &=& \psi\left(\frac{\nabla \rho}{\rho}+ik\nabla S\right)\\ \nabla^2 \psi &=& \psi\left(\frac{\nabla^2 \rho}{\rho}-\left(\frac{\nabla \rho}{\rho}\right)^2+ik\nabla^2 S\right)+\psi\left(\frac{\nabla \rho}{\rho}+ik\nabla S\right)^2 \end{eqnarray} and obtain the equations \begin{eqnarray}\frac{\nabla^2 \rho}{\rho} - k^2 \left(\nabla S\right)^2 +k^2 n^2&=&0\,, \\\rho \nabla ^2S + 2\nabla \rho \cdot \nabla S&=& 0\,.\end{eqnarray}

When $\lambda \rightarrow 0$, $k\rightarrow\infty$, in the first equation, the second and third terms dominate, leading to the constraint \begin{equation} \boxed{(\nabla S)^2 = n^2}\,.\tag{4}\end{equation} This is the eikonal equation.

The surface $S(\vec{r})=\text{constant}$ is the wavefront, and its gradient $\nabla S$ is ray in geometrical optics. Consider the motion along the ray, we have the relation \begin{equation}\frac{dr}{ds}=\frac{\nabla S}{\left|\nabla S\right|}=\frac{\nabla S}{n}\,,\end{equation} where we use the eikonal equation (4) in the second step. Taking the total derivative with respect to $s$ on both sides of the equation \begin{equation} \nabla S = n\frac{dr}{ds}\,,\end{equation} the R.H.S is simply $\frac{d}{ds}\left(n\frac{dr}{ds}\right)$ and the L.H.S. is \begin{equation}\frac{d}{ds}\nabla S = \left(\frac{d\vec{r}}{ds}\cdot \nabla\right)\nabla S= \left(\frac{\nabla S}{n}\cdot \nabla\right)\nabla S=\frac{1}{2n}\nabla\left(\nabla S \right)^2 = \frac{1}{2n}\nabla n^2=\nabla n\,.\end{equation} The final result is exactly Eq. (2).

Classical limit of Schrodinger equation

In the quantum mechanics, the wave function $\psi(x, t)$ is described by the Schrodinger equation \begin{equation}i\hbar\frac{\partial}{\partial t}\psi(x,t)= \left[-\frac{\hbar^2}{2m} + V(x)\right]\psi(x, t)\,.\end{equation}

The quantum mechanics should become the classical mechanics in the limit $\hbar\rightarrow 0$. The proof is in two key steps:

1. Consider the ansatz of the waveefunction \begin{equation}\psi(x,t)=\rho(x,t)e^{\frac{i}{\hbar}S(x,t)}\,.\end{equation} and plug it into the Schrodinger equation.
2. Show that when $\hbar\rightarrow 0$, one of the equations on $S(x, t)$ becomes \begin{equation}\frac{\partial S}{\partial t}+\frac{1}{2m}(\nabla S)^2 +V=0\,,\end{equation} which is the Hamilton-Jacobi equation in the classical mechanics.

The ansatz in the first step is inspired by the transition from wave optics to geometrical optics at the limit of zero wavelength as described in the previous section, in which we know that it is the phase of a wave that matters at the geometrical limit. Moreover, the result in the second steps reveals that the phase $S(x,t)$ of a wavefunction $\psi(x, t)$ is the Hamilton's principal function in the classical limit. I have another post to explain what the Hamilton-Jacobi equation is and how it describes the classical motion.