Mathematical proof regarding angle mirroring

83 Views Asked by At

I'm trying to find a proof for a statement that is made in Griffiths' Introduction to Electrodynamics. It can be stated as follows (in my own words):

Let $P$ be a point such that its angle in polar coordinates is given by $\theta>0$. Let $\beta > \theta$ be the angle between any two lines in the 2D plane. Then, if we try to find all the points such that both lines act as mirrors, if $\beta$ is an integer divisor of $180$, all of these points will fall outside the region delimited by the angular interval $(0, \beta)$, independently of the value of $\theta\in (0,\beta)$.

Let me show some examples to make this clearer.

Example 1

$\beta = 45^\circ$, $\theta = 30^\circ$

enter image description here

The red dot represents the original $P$ point. Note that the blue points are placed in a way that make the lines with slope $0$ and $\beta$ 'mirrors'. Due to the fact that $\beta = \frac{180^\circ}{4}$, no blue dots appear inside the region delimited by the black lines.

Example 2

$\beta = 50^\circ$, $\theta = 15^\circ$

enter image description here

In this case, due to the fact that $\frac{180^\circ}{\beta}$ is not an integer, blue dots appear inside the region delimited by the black lines. This is, however, not general. It is possible to find pairs of values of $\theta$ and $\beta$ such that no blue dots appear in the region, but it is not likely.

I've tried with some values and the statement made by Griffiths seems to be valid. Is there a way to prove this generically?

1

There are 1 best solutions below

1
On BEST ANSWER

Let $\ell_1$ and $\ell_2$ be the two mirror lines, and assume $\beta={\pi\over n}$. Denote the reflections in the $\ell_i$ by $\rho_i$. You then obtain blue points from the starting red point $P$ by applying alternatively $\rho_1$ and $\rho_2$. Now it is easy to see that $\rho_2\circ\rho_1$ is a rotation by the angle $\alpha=|2\beta|={2\pi\over n}$ in one of the two directions. It follows that $(\rho_2\circ\rho_1)^n$ is the identity, hence gives back $P$. In all you obtain the $2n$ points $$(\rho_2\circ\rho_1)^k (P)\quad(0\leq k\leq n-1),\qquad (\rho_2\circ\rho_1)^k\circ\rho_2 (P)\quad\quad(0\leq k\leq n-1)\ ,$$ one in each sector of central angle $\beta={2\pi\over 2n}$.