# Lecture 10: Governing principles, Optics

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## Prelude

• Discovery of 9th planet (Bacchus, Bowie)
• Vulcan and Mercury and Einstein's explanation with general relativity
• Presentations from last time
• Email a pdf of your presentation by 10:30, Friday morning.

## Introduction

So far, we've discussed using compartmental models translated into differential equations with the law of mass action and the principle of conservation of mass to describe population interactions. But differential equations arrise in many contexts as descriptions of phenomena and consequences of general organizing principles.

Organizing principles (super-models)

• Aristotle -- the world is made as what is "best"! (all beautiful things are useful)
• path of least time
• Conservation of energy
• Minimum potential energy
• Newton's laws

## A focusing mirror

Fermat's principle: "Light follows the path of least time between two points."
Today, based on our modern understanding of light as part wave, we say instead that light follows paths of stationary action (which might be a minimum or maximum of the distance travelled).

Optics: Suppose we would like to build a mirror that reflects parallel light rays to a single focal point.What shape should that be?

Take the focus at $$(0,0)$$. Start a curve $$y(x)$$ at $$y(0) = -q$$. At every point, $$(x, y(x))$$. The basic law of optics says that when light bounces off of a mirror, the angle of incidence equals the angle of reflection.

The reflection matrix $\mathrm{R} = \mathrm{I} - 2 \frac{v v^T}{v^T v}$ is such that the part perpendicular to v stays the same, but the part parallel to v is reflected. We can show that this matrix does what we say it will by breaking any vector x up into parts parallel to v and orthogonal to v.

Tangent vector $$(1, dy/dx)$$. Let's define the normal vector $v= (-dy/dx, 1)$.

If we take a ray from our focus and bounce it off of our mirror, we want to reflected ray $$(x, y(x))$$ to be vertical, hence perpendicular to $$(1,0)$$. So, $\begin{bmatrix} 1 & 0\end{bmatrix} \mathrm{R} \begin{bmatrix} x\\y\end{bmatrix} = 0,$ $\begin{bmatrix} 1 & 0\end{bmatrix} \left( \mathrm{I} - 2 \frac{v v^T}{v^T v} \right) \begin{bmatrix} x\\y\end{bmatrix} = 0,$ $\begin{bmatrix} 1 & 0\end{bmatrix} \left( \begin{bmatrix} 1&0\\0&1 \end{bmatrix} - \frac{1}{y'^2 + 1} \begin{bmatrix} y'^2 & -y'\\-y'& 1 \end{bmatrix} \right) \begin{bmatrix} x\\y\end{bmatrix} = 0.$

We can perform the calculation using the sympy package for symbolic compuations in python

v = Matrix([[-y(x).diff(x)],[1]])
R = 2 * (v * v.T)/(v.T * v)[0,0] - eye(2)
(Matrix([[1,0]]) * R * Matrix([[-x],[q-y(x)]]))[0,0].factor()

The matrices reduce to the scalar nonlinear ordinary differential equation $0 = -\frac{ \left(x \left[ \left(\frac{d y}{d x} \right)^{2} - 1\right] - 2 y(x) \frac{d y}{d x} \right) }{\left(\frac{dy}{dx} \right)^{2} + 1}$

Just from looking at this equation, we can learn things. The equation is first order, nonlinear, and quadratic in the first-derivative term. It is a kind of equation we never see in standard classes on differential equations, and so fancy usual solution methods won't be of much help. Thus, if we know a position $$(x,y(x))$$ where the mirror passes, we can determine the slope of the curve at that point by using the quadratic formula. But, the equation has 2 solutons for the slope most of the time! Why? Well, a little thought reveals that this makes perfect sense -- there are actually two different mirrors that solve our equations -- one that collects light from above, and one that collects light from below! So, this ambiguity actually makes good sense.

This is a highly nonlinear first-order equation without unique solutions, BUT if we try an ansatz $$y(x) = a x^2 + b$$, we find this is a solution for any values of $$a$$ and $$b$$ when $$4ab+1=0$$ $y(x) = b - \frac{x^2}{4b}$ So, if the parabola passes below the focus ($$b < 0$$), then it is convex, and if the parabola passes above the focus ($$b > 0$$), then it is concave, and the equ