Rotating Mirror

You can do interesting things with mirrors. You can find out for yourself with the “rotating mirror” and “funnel” exhibits. Mirrors have fascinated mankind since the Stone Age. But what is the mathematics behind it?

And now … the mathematics:

Mathematically speaking, a reflection on a plane simply consists of inverting an axis of a rectangular coordinate system. For example, a reflection on the $xy$ -plane (for example, a water surface) is completely represented by the mapping $s$ given by equation

$s\Biggl(\begin{pmatrix} x\\ y\\ z\end{pmatrix}\Biggr)=\begin{pmatrix} x\\ y\\ -z\end{pmatrix}=\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & -1\end{pmatrix}\begin{pmatrix} x\\ y\\ z\end{pmatrix},$

described. The preimage (for example, the sky) is thereby mapped onto its mirror image (the sky suddenly appears to be under the surface of the water). What is striking is that the orientation is reversed: if you hold your right hand in the mirror, for example, your reflection raises your left hand.

But what happens now if we move and rotate the mirror plane $E$ in space? Say this passes through the point $\mathbf{p}$ and has the normal vector $\mathbf{n}$ . We now want to determine how the associated reflection $s_{\mathbf{p},\mathbf{n}}$ maps any point $\mathbf{x}$ . To do this, we first form the perpendicular from $\mathbf{x}$ to $E$ (i.e. the line whose one vertex is $\mathbf{x}$ and whose other vertex lies in $E$ and is perpendicular to $E$ ). This has exactly the length $l=\langle\mathbf x-\mathbf p,\mathbf n\rangle$ , where $\langle\mathbf u,\mathbf v\rangle$ is the scalar product of the vectors $\mathbf u$ and $\mathbf v$ . This is exactly the projection of the line $\overline{\mathbf x\mathbf p}$ onto the normal vector $\mathbf n$ . The mirror image of the point $\mathbf x$ on the plane $E$ is now exactly the corner point of the mirrored perpendicular not lying in $E$ . So we have to subtract the distance $l$ twice in the direction of the normal vector of $\mathbf x$ . So the equation

$s_{\mathbf p,\mathbf n}(\mathbf x)=\mathbf x-2\langle\mathbf x-\mathbf p,\mathbf n\rangle\mathbf n$

results.

To the “Revolving Mirror” exhibit

But now we turn our attention to the “rotating mirror” exhibit. You are probably familiar enough with the case where there is only a mirror in front of you in everyday life. You simply see yourself mirrored in him. If you turn the mirror, nothing changes at all, because the mirror plane remains the same.

But now let’s start from the more interesting case where there are two mirrors (planes) $E_1$ and $E_2$ that have the normal vectors $\mathbf n_1$ and $\mathbf n_2$ and intersect at the point $\mathbf p$ . Now what happens when you look into such a construction? We can easily deduce (“work out”) this with the above considerations: Let $s_1$ and $s_2$ be the reflections on the plane $E_1$ and $E_2$ , respectively. We get

$\begin{gather*}s_1(s_2(\mathbf x))=s_1(\mathbf x-2\langle\mathbf x-\mathbf p,\mathbf n_2\rangle\mathbf n_2)=x-2\langle\mathbf x-\mathbf p,\mathbf n_2\rangle\mathbf n_2-2\langle x-2\langle\mathbf x-\mathbf p,\mathbf n_2\rangle\mathbf n_2-\mathbf p,\mathbf n_1\rangle\mathbf n_1\\ =\mathbf x-2(\langle\mathbf x-\mathbf p,\mathbf n_1\rangle\mathbf n_1+\langle\mathbf x-\mathbf p,\mathbf n_2\rangle\mathbf n_2)+4\langle\mathbf x-\mathbf p,\mathbf n_2\rangle\langle\mathbf n_2,\mathbf n_1\rangle\mathbf n_1\quad(\ast).\end{gather*}$

Here we have used the bilinearity of the scalar product. A number of things can be read from this expression: For example, it is generally not symmetric in $\mathbf n_1$ and $\mathbf n_2$ , i.e. does not coincide with the double mirror image $s_2(s_1(\mathbf x))$ . Symmetry (and thus equality of the two expressions) occurs exactly when the last summand in the above equation $(\ast)$ becomes zero, i.e. $\langle\mathbf n_1,\mathbf n_2\rangle=0$ , i.e. $E_1$ and $E_2$ are perpendicular to each other (intersection angle $\alpha=90^\circ$ ). So in this case it doesn’t matter which of the two mirrors you look into — you don’t see a break at the intersection line. You can check this yourself on the exhibit: With the rotating mirror, where both mirror planes meet perpendicularly, there is no “break” at the intersection line. With the other one, however, it is. This corresponds exactly to the above observation.

But what is actually happening at the two mirrors? Let $g=E_1\cap E_2$ be the intersection of the mirror planes $E_1$ and $E_2$ . We look at the whole construction from “above“, i.e. along $g$ . To do this, we rotate our coordinate system so that $E_1$ just becomes the $xz$ -plane and $g$ becomes the $z$ -axis. Then the projection of $E_2$ along the straight line $g$ onto the $xy$ -plane becomes exactly an origin line intersecting the positive $x$ -axis at the intersection angle $\alpha=\arccos(-\langle\mathbf n_1,\mathbf n_2\rangle)$ . In these new more suitable coordinates we can now easily illustrate what happens to a point $\mathbf x=(x,y,z)$ . At $s_1$ this is mapped to the point $\mathbf x_1=s_1(\mathbf x)=(x,-y,z)$ , which then maps under $s_2$ to the point

$\mathbf x_2=s_2(\mathbf x_1)=\begin{pmatrix}\cos(2\alpha)x-\sin(2\alpha)y\\ \sin(2\alpha)x+\cos(2\alpha)y\\ z\end{pmatrix}$

goes. See also figure 1 below:

It is therefore simply a rotation by the angle $2\alpha$ around the $z$ axis. In the same way it can be determined that thus the point $\mathbf x_2'=s_1(s_2(\mathbf x))$ is simply the point $\mathbf x$ rotated by $-2\alpha$ about the $z$ -axis. This in turn also confirms our above observation that the mappings $s_1\circ s_2$ and $s_2\circ s_1$ are exactly equal for $\alpha=90^\circ=\pi/2$ , because then both are simply equal to a reflection on the straight line $g$ !

This now even explains the observation that the image you see in the two rotating mirrors with two mirror planes rotates when you set the construction in rotation. Because the intersection line $g$ then rotates in front of you and thus also the mirror images.

Three and more mirrors

If you now add another mirror, it gets even more curious: Let’s assume that three mirrors with the mirror planes $E_1$ , $E_2$ , $E_3$ are perpendicular to each other (thus forming the coordinate planes $xy$ , $xz$ and $yz$ except for rotation). Similar considerations as above, now show that the threefold mirrored point $\mathbf x$ is then transformed into the point $-\mathbf x$ (independent of the order of the mirrorings; see figure 2). Similar considerations as above, now show that the threefold mirrored point $\mathbf x$ is then transformed into the point $-\mathbf x$ (independent of the order of the mirrorings; see figure 2). It gets even better: No matter from which direction you look into this construction, you always see your face, because the point $\mathbf x$ is always exactly opposite the triple mirror image $-\mathbf x$ . This technique is also used in shipping, for example for bridges.

Mirror groups

If you look into two mirrors, as in the rotating mirror exhibit, you may have noticed that from some viewpoints it appears as if not just two, but several mirrors are at the same angle to each other. How many mirrors you see depends on the intersection angle $\alpha$ . If, for example, the two mirrors meet at an angle of $90^\circ$ , it seems to you that there are four mirrors evenly arranged around the intersection line. With a smaller cutting angle, the number becomes larger. So where does this strange phenomenon come from?

Now, above we have seen that the order in which the reflections take place is important. Thus, running $s_1\circ s_2$ and $s_2\circ s_1$ one after the other results in a different image (at least if the intersection angle $\alpha$ is not exactly $90^\circ$ ).

What happens now is that you see not only the mirror image of the mirror image, but the mirror image of the mirror image of the mirror image and so on. This means on the mathematical side that you determine all possible mappings $s_1\circ s_2\circ s_1\circ\cdots$ that can somehow be composed of the reflections $s_1$ and $s_2$ . This is called the group generated by the reflections $s_1$ and $s_2$ . A mirror image always has the property that it produces the identical image when applied twice: if you change the sign of a basis vector of an orthonormal basis twice, you get the original basis back. So that means $s_1\circ s_1=s_2\circ s_2=\mathrm{id}$ . We have also already considered that $s_1\circ s_2$ represents a rotation by the angle $2\alpha$ (and $s_2\circ s_1$ a rotation in the opposite direction). This makes it appear to you that many mirrors are arranged around the straight line $g$ so that two intersect each other at an angle of $\alpha$ (because the plane $E_1$ then intersects with the mirror image $s_1(E_2)$ exactly at the angle $\alpha$ ).

The associated group is called a dieder group. These are groups generated by exactly two reflections (these are also called involutions, i.e. $s^2=\mathrm{id}$ ). How many elements the dieder group now has depends on the number of different mappings that can be written as concatenations of the two basic mappings. This in turn depends on the angle $\alpha$ : For example, if $\alpha=90^\circ$ , then $s_1$ and $s_2$ interchange so that $s_1\circ s_2=s_2\circ s_1$ . It is then easy to consider that there are only four fundamentally different mappings: $\mathrm{id},\s_1,s_2,s_1s_2$ . This corresponds to the dieder group $D_2$ of order $4$ . This is the symmetry group of a line in the plane. Is now $\alpha=2\pi\frac{p}{q}$ is a rational multiple of the total angle $2\pi$ with $p,q\in\mathbb N_+$ divisible, then the group generated by $s_1$ and $s_2$ is the dieder group $D_q$ of order $2q$ , i.e. the transformation group of a regular $q$ -corner, because $(s_1\circ s_2)^q=\mathrm{id}$ . If, on the other hand, $\alpha$ is not such a rational multiple of $\pi$ , then the rotation $s_1\circ s_2$ never returns to its initial state, i.e. it does not satisfy an equation of the form $(s_1\circ s_2)^q=\id$ . This gives us the infinite dieder group $D_\infty$ .

For two mirrors, that’s all that can happen. If, on the other hand, you take three mirrors or more, it becomes more complicated: With three mirrors standing vertically on top of each other, you get a group with eight elements, each of which transforms the unit cube into itself.

Such groups, which are generated by finitely many reflections, can be studied and classified in detail, see [1].

The exhibits “Kaleidoscope Mirror“, “Kaleidoscope“, “Mirror Funnel” and “Polyhedron Crown” are also relevant to this. The first three again show a group created by a certain arrangement of mirrors. The “mirror funnel” is particularly interesting here, because it seems as if one is seeing the sides of a dodecahedron here. This connection is no coincidence, for a Platonic solid is transformed into itself by every reflection on a plane that passes through its centre and contains one of its edges.

Literature

[1] https://en.wikipedia.org/wiki/Root_system

[2] https://en.wikipedia.org/wiki/Dihedral_group

[3] https://en.wikipedia.org/wiki/Regular_polygon

[4] https://en.wikipedia.org/wiki/Platonic_solid