Analytical Optics

July 2023

The calculus of variations provides a very nice abstraction barrier for studying nature. Step one, find a variational principle for the phenomena you're studying. Step two, use conservation laws, generalized force, momentum, energy, etc. to make some deductions. Lagrange and Hamilton already did this to classical mechanics (I've got two posts about that) and Fermat found a variational principle for optics back in the 17th century. A few months ago, when I finised by post about geometrical optics I thought of joining in the fun and finishing what Fermat started by doing some optics analytically. Watch me use no figures in this entire post.

Fermat's mathematical superpower was definitely his skill of asking "what if?" appropriately. He was looking at Pythagorean triples, "what if the power isn't two?" Later on he was reading an old optics book that said that the movement of light in a uniform medium can be described as the path of least length, "well that doesn't work for non-uniform media, what if its the path of least time?" He also knew some basic calculus by then so he got to work and managed to prove Snell's law this way. In short, he found a variational principle for optics.

Axiom 1. The path $\gamma : [0,1] \to \R^2$ taken by light is the path that satisfies $\delta T = 0$ where $T$ is the time it takes light to traverse $\gamma.$ This is often called Fermat's principle or the principle of least time.

Suppose that $\gamma(t) = (x(t), y(t)).$ The time it takes for light to traverse an infinity small part $\d s$ of $\gamma$ is $\d T = \d s/v(x,y)$ where $v(x,y)$ is used to indicate the speed of light at the point $(x,y).$ It is clear that $\d s = \sqrt{\d x^2 + \d y^2}.$ Since $\d x = \dot{x}\,\d t$ and $\d y = \dot{y}\,\d t,$ then $\d s = \sqrt{\dot{x}^2 + \dot{y}^2}\,\d t.$ Restating $\d T$ in this way results in $\d T = \d t\sqrt{\dot{x}^2 + \dot{y}^2}/v(x,y).$ Therefore the time it takes for light to traverse $\gamma$ is $$ \begin{equation} T = \int_0^1 \frac{\sqrt{\dot{x}^2 + \dot{y}^2}}{v(x,y)}\,\d t. \end{equation} $$ This integral is clearly analogous to the action of classical mechanics. The integrand is also analogous to the Lagrangian. So we will be calling $$\begin{equation} L = \frac{\sqrt{\dot{x}^2 + \dot{y}^2}}{v(x,y)} \end{equation}$$ the optical Lagrangian from now on. We have now officially crossed the abstraction barrier of the variational calculus, and now can use its results to make some deductions.

The greatest result of the calculus of variations is the Euler-Lagrange equation. $$ \begin{equation} \frac{\d}{\d t}\frac{\partial L}{\partial \dot{q}} = \frac{\partial L}{\partial q}. \end{equation} $$ Where $\partial L/\partial \dot{q}$ is the generalized momentum and $\partial L/\partial q$ is the generalized force. We may find the momentum and force in an optical system by substituting (2) to attain $$ \begin{align} \vec{p} &= \left(\frac{\dot{x}}{v\sqrt{\dot{x}^2 + \dot{y}^2}}, \frac{\dot{y}}{v\sqrt{\dot{x}^2 + \dot{y}^2}}\right), \\ \vec{F} &= \left(\frac{-\sqrt{\dot{x}^2 + \dot{y}^2}}{v^2}\frac{\partial v}{\partial x}, \frac{-\sqrt{\dot{x}^2 + \dot{y}^2}}{v^2}\frac{\partial v}{\partial y} \right). \end{align} $$ Using these expressions and the fact that $\vec{F} = \dot{\vec{p}}$ we can find the differential equations for the motion of light. $$ \begin{aligned} \frac{(\dot{x}^2+\dot{y}^2)(\ddot{x}v-\dot{x}\dot{v}) - \dot{x}v(\dot{x}\ddot{x}+\dot{y}\ddot{y})} {v^2(\dot{x}^2+\dot{y}^2)^{3/2}} &= \frac{-\sqrt{\dot{x}^2 + \dot{y}^2}}{v^2}\frac{\partial v}{\partial x}, \\ \frac{(\dot{x}^2+\dot{y}^2)(\ddot{y}v-\dot{y}\dot{v}) - \dot{y}v(\dot{x}\ddot{x}+\dot{y}\ddot{y})} {v^2(\dot{x}^2+\dot{y}^2)^{3/2}} &= \frac{-\sqrt{\dot{x}^2 + \dot{y}^2}}{v^2}\frac{\partial v}{\partial y}. \end{aligned} $$ Yikes. If you're brave enough you can try to do something with this, but this ain't for me. Especially because—as we'll see shortly—$v(x,y)$ is almost never differentiable in a real world scenario.

For example, when we're consider light's traversal through the boundary of two different media, we encounter a scenario where $v$ is not differentiable.

Suppose that the line $x = 0$ is the boundary of two regions of space. To the left of $x=0$ the speed of light is $v_1$ and to the right of it the speed of light is $v_2.$ Therefore $$ v(x) = \begin{cases}v_0 & x < 0 \\ v_1 & x > 0 \end{cases} $$ which is obviously not differentiable. That the optical Lagrangian $L$ given by this choice of the speed of light has symmetry in the $y$-axis. That is, if $\Delta y$ is any constant $$\begin{aligned} L(x, y + \Delta y, \dot{x}, \dot{y}) = \frac{\sqrt{\dot{x}^2 + \dot{y}^2}}{v(x, y + \Delta y)} = \frac{\sqrt{\dot{x}^2 + \dot{y}^2}}{v(x, y)} = L \end{aligned}$$ This means that $L$ is constant with respect to $y$. Using this and the Euler-Lagrange equation we can see that $$ \frac{d}{dt}\frac{\partial L}{\partial\dot{y}} = \frac{\partial L}{\partial y} = 0 $$ so that the second component of generalized momentum, $\partial L/\partial\dot{y}$ is a conserved quantity; which we could have also seen by Noether's theorem. This means that if $(x_0, y_0)$ and $(x_1, y_1)$ are on opposite sides of the $y$-axis, then by (4) $$ \frac{\dot{y}_0}{v_0\sqrt{\dot{x}_0^2 + \dot{y}_0^2}} = \frac{\dot{y}_1}{v_1\sqrt{\dot{x}_1^2 + \dot{y}_1^2}} $$ Now, notice that $\dot{y} / \sqrt{\dot{x}^2 + \dot{y}^2}$ is equal to the sine of the angle $\theta$ made by the tangent vector $(\dot{x},\dot{y})$ and the $x$-axis. At $(x_0, y_0)$ we denote this angle by $\theta_0$ and call it the angle of incidence. Similarly, at $(x_1, y_1)$ we denote it by $\theta_1$ and call it the angle of refraction. We conclude that $$ \frac{\sin\theta_0}{v_0} = \frac{\sin\theta_1}{v_1} $$ which is the widely known Snell's law of refraction.

The law of reflection follows as a special case of the law of refraction. If the $y$-axis is a mirror instead of the boundary between two media, then $v_0 = v_1$ implying that $\theta_0 = \theta_1$ Therefore, the angle of incidence is equal to the angle of reflection (which takes the place of the angle of refraction).

Theorem 1. When light bounces off of a mirror, the angle of incidence is equal to the angle of reflection.

And that's about it for this very theory heavy post. No new results, all of it just a fancy restatement of what we knew in the last post. To compensate I'll make a post about some of the nice exercises in A Treatise on Geometrical Optics sometime in the near future.

1. The variable $t$ isn't really time. It's just a parameter with which we define the path of light parametrically. I would've used another symbol like $\tau$ if it wasn't hard to type repeatedly.