Spin-1/2 and a rotating mirror

Difficulty:   ★★☆☆☆   high school

Imagine a light ray reflecting off a mirror. If the mirror is rotating, the direction of the reflected beam will also rotate, but at twice the rate of the mirror! This follows from the way the angles work, if you recall for example “the angle of incidence equals the angle of reflection” and think about it carefully… Or, just play around with an animation until it looks right 😉 . In quantum physics this angle-doubling property turns up in the description of electrons for example, where it seems very mysterious and exotic (keywords: “spin-1/2” and “spinors”). So its appearance in an “ordinary” and intuitive setting is reassuring.

angles for the mirror and ray
Figure 1: A two-sided mirror and light ray (green arrows). The light arrives from the left, bounces off the mirror in the centre, and exits towards the upper-right. In this example the light beam has angle b = 165°, and the mirror 15°.

For simplicity let’s use a two-dimensional plane, as shown in Figure 1. We measure angles from the positive direction of the x-axis, as usual in polar coordinates. I choose to measure from the centre outwards, so for the incoming ray the angle assigned is the opposite of what the arrow might suggest. Label the incoming ray angle b, and mirror rotation angle m. Now if you increase m by some given amount, the outgoing ray angle increases by twice as much. But if you increase b instead, the outgoing ray angle decreases by the same amount. We also need an “initial condition” of sorts:  when the mirror is horizontal (m = 0°), and the ray arrives from directly above (b = 90°), the reflected beam is also at 90°. It follows:

reflected angle  =  2mb + 180°.

Now if the mirror rotates by 180°, the reflected ray completes a full 360° rotation, so is back to its original position. (We suppose the mirror is 2-sided.) If you hadn’t watched the rotation, you wouldn’t know anything had changed. But now suppose we make one side of the mirror red and the other blue, so the reflected ray takes on the colour of the closest side. Now the ray must make two complete revolutions, 720°, to get back to its original state! After one revolution it is back to the same position, but has a different colour, as the opposite side of the mirror faces the beam. Similarly, if the reflected ray is rotated by 180° in one direction, this is not the same as rotating by 180° in the opposite direction, as the colour is different. “Spinors” have these same features, except in place of red/blue their mathematical description picks up a factor of ±1.

two mirrors rotated by differing amounts
Figure 2: A two-sided coloured mirror. The incoming ray is depicted as grey, but think of it as white so its reflection is free to take on the appropriate colour. In the left image, the blue side of the mirror is facing upwards. Now slowly rotate the mirror by a half-revolution. (Actually I have drawn a slightly different angle just for variety.) The reflected ray changes angle by a full revolution, but has since turned red!

You might try animating this yourself. If you draw the rays with unit length, then the arrow for the incoming beam points from (cos b, sin b) to (0,0). The outgoing arrow points from (0,0) to -(cos(2mb), sin(2mb)), where a minus sign replaces the 180° term from earlier. The colour depends on whether the incoming ray is from 0 to 180° ahead of the mirror, or 0 to 180° behind. This is determined by the sign of sin(mb). It is convenient to allow angle parameters beyond 360°, which makes  no physical difference  at most only a change in colour, as we have learned 😀 . Below is Mathematica code I wrote, which uses slider controls for the angle parameters. The result is fun to play around with, and it helps make the angle-doubling more intuitive.

width = 0.3;
height = 0.02;
mirror = Polygon[{{-width,-height},{width,-height},{width,height},{-width,height}}, VertexColors->{Red,Red,Blue,Blue}];
Manipulate[ Graphics[ {Rotate[mirror,m], {Gray,Arrow[{{Cos[b],Sin[b]},{0,0}}]}, {If[Sin[b-m]>0,Blue,Red],Arrow[{{0,0},-{Cos[2m-b],Sin[2m-b]}}]}}, PlotRange->{{-1,1},{-1,1}} ], {{m,0},-2\[Pi],2\[Pi]}, {{b,3\[Pi]/4},0,2\[Pi]} ]

Leave a Reply

Your email address will not be published. Required fields are marked *