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]} ]

More on “covariant” versus “contravariant”

Difficulty:   ★★★☆☆   undergraduate

I have written on the topic of “covariant” and “contravariant” vectors (and higher-rank tensors) previously, and have been intending to write an update for a number of years. It must be noted some authors recommend avoiding these terms completely, including Schutz 2009  §3.3:

Most of these names are old-fashioned; ‘vectors’ and ‘dual vectors’ or ‘one-forms’ are the modern names.

Let’s return to Schutz’ reason soon. I have followed this naming practice myself, except I prefer to say “covector” or “1-form”, rather than “dual vector” which can be clumsy. (What would you call the vector which is the (metric) dual to a given 1-form: the “dual of a dual vector”, or a “dual-dual-vector”!?) People also talk about “up[stairs] indices” and “down[stairs] indices”, which seems alright.

But if you want to be cheeky, you might say a vector is covariant, while a 1-form is contravariant — the exact opposite of usual terminology! I remember a maths graduate student at some online school stating this. Similar sentiments are expressed by Spivak 1999 vol. 1  §4:

Nowadays such situations are always distinguished by calling the things which go in the same direction “covariant” and the things which go in the opposite direction “contravariant”. Classical terminology used these same words, and it just happens to have reversed this: a vector field is called a contravariant vector field, while a section of T*M is called a covariant vector field. And no one has had the gall or authority to reverse terminology so sanctified by years of usage. So it’s very easy to remember which kind of vector field is covariant, and which is contravariant — it’s just the opposite of what it logically ought to be.

While I love material which challenges my conceptual understanding, and Spivak’s humorous prose is fun; trying to be too “clever” with terms can hamper clear communication. Back to Schutz, who clarifies:

The reason that ‘co’ and ‘contra’ have been abandoned is that they mix up two very different things: the transformation of a basis is the expression of new vectors in terms of old ones; the transformation of components is the expression of the same object in terms of the new basis.

If you take the components of a fixed vector in a given basis, they transform contravariantly when the basis changes. But if you consider the vector as a whole — a single geometric object — and ask how a basis vector (specifically) is mapped to a new basis vector, the change is “covariant”. (Recall, as Schutz explains: “The property of transforming with basis vectors gives rise to the co in ‘covariant vector’ and its shorter form ‘covector’.”) In general, if you fix a set of components, by which I mean fixing an ordered set of numbers like (0,1,0,½) say, and then change the basis vectors these numbers refer to, then the change of a vector (as a whole entity) is “covariant”, so-called. For 1-forms, the converse of these statements apply. Some diagrams would make this paragraph clearer, but I leave this as an exercise, sorry.

However, it seems to me the most accurate description is that vectors don’t change at all, when you change a basis! Picture a vector as an arrow in space, then the arrow does not move. In this sense vectors are neither contravariant nor covariant, but invariant! (We could also say generally covariant, since they are geometric entities independent of any coordinate system. This is the usual modern meaning of the word “covariant”, but it’s a bit different to the covariant–contravariant distinction, so for clarity I avoid this language here.) In conclusion, one of the clearest descriptions is to simply say: vector, or 1-form / covector. Or, given historical usage, it is especially clear to say a vector’s “components transform contravariantly”. See the Table.

Table: transformation under basis change
object clarification vector 1-form
components same (co)vector, but components in a new basis contravariant covariant
(co)vector same (co)vector, treated as a whole invariant invariant
basis (co)vector transform to different (co)vector covariant contravariant

Addendum: I recently learned a dual basis need not be made up of 1-forms, as in the usual formulation in differential geometry, but of vectors instead! Recall the defining relation between a coordinate basis and cobasis: dx^\mu(\partial_\nu) = \delta^\mu_\nu, or \mathbf e^\mu(\mathbf e_\nu) = \delta^\mu_\nu for an arbitrary frame. In particular, each cobasis element is orthogonal to the “other” 3 vectors. But we can take the duals (dx^\mu)^\sharp, which are vectors but obey the same orthogonality relations, via the metric scalar product: \langle(dx^\mu)^\sharp,\partial_\nu\rangle = \delta^\mu_\nu. (By relating to the standard approach I may have made things look complicated, but this should be visualised as simply finding new vectors orthogonal to existing vectors.) (On a separate note, “dual” here means as an individual vector, not dual as a basis.) It seems this vector approach to a dual basis was the original one. In 1820 an Italian mathematician Giorgini distinguished between projezione oblique (parallel projections) and projezioni ortogonali (orthogonal projections) of line segments, now termed contravariant and covariant. According to one historian (Caparrini 2003 ), this was “one of the first clear-cut distinctions between the two types of projections in analytic geometry.” However priority goes to Hachette 1809 . Today in geometric algebra, also known as Clifford algebra, a dual basis is also defined as vectors not 1-forms, via \langle\mathbf e^\mu,\mathbf e_\nu\rangle = \delta^\mu_\nu (Doran & Lasenby 2003  §4.3).