Adventures in Parallel Dimension #1729 IV

One of the defining characteristics of light is the principle of least time, which states that a light beam will always take the path from A to B which minimizes the travel time.

We can express the travel time $T$ as

$T = \int_{A}^{B} \frac{dl}{v} = \frac{1}{c} \int_{A}^{B} n(y) \sqrt{dx^{2}+dy^{2}} = \frac{1}{c} \int_{A}^{B} n(y) \sqrt{1+\dot{x}^{2}}dy$

Where $A,B$ are fixed points and $\dot{x}$ is the derivative of $x$ with respect to $y$ . We can then write this as a Lagrangian

$T = \frac{1}{c} \int_{A}^{B} \sqrt{y}\sqrt{1+\dot{x}^{2}}dy = \frac{1}{c} \int_{A}^{B} L(x,\dot{x},y)dy$

Where $L(x,\dot{x},y) = \sqrt{y}\sqrt{1+\dot{x}^{2}}$ . This is then extremized whenever $L$ satisfies the Euler-Lagrange equation

$\frac{\partial L}{\partial x} = \frac{d}{dy} \frac{\partial L}{\partial \dot{x}} \Rightarrow 0 = \frac{d}{dy} \frac{\dot{x}\sqrt{y}}{\sqrt{1+\dot{x}^{2}}} \Rightarrow \frac{\dot{x}\sqrt{y}}{\sqrt{1+\dot{x}^{2}}} = C$

Where $C$ is a constant defined by our initial conditions. It follows that

$C^{2}+C^{2}\dot{x}^{2}=\dot{x}^{2}y \Rightarrow dx = \frac{C}{\sqrt{y-C^{2}}}dy \Rightarrow \int dx = \int \frac{C}{\sqrt{y-C^{2}}}dy$

After integrating both sides

$x+D = 2C\sqrt{y-C^{2}} \Rightarrow \frac{(x+D)^{2}}{4C^{2}}+C^{2}=y$

Where $D$ is another constant defined by boundary conditions. Since $y(a) = y(-a)$ , it follows that $y$ is even and thus $D=0$ . We then solve for $C^{2}$

$\frac{a^{2}}{4C^{2}}+C^{2} = a \Rightarrow (2C^{2}-a)^{2} = 0 \Rightarrow C^{2} = \frac{a}{2}$

And so $y(x)$ becomes

$y(x) = \frac{x^{2}}{2a}+\frac{a}{2}$

Which has a minimum value of $y=\frac{a}{2}$ at $x=0$ . Hence, the minimum value of $y$ occurs at $(0,\frac{a}{2})$ , and the answer is $\boxed{50}$