Dust circles

The subject of the exhibit “Dust Circles” is an interesting mathematical theorem about plane motions. First of all you can look at the exhibit: It consists of two transparent plastic plates, each of which has black dust embedded in a congruent manner. The upper of the two is clamped in a wooden frame, which can be moved against the lower fixed plate. What can you observe now?

It seems that circles are involuntarily formed by this displacement — as the name of the exhibit suggests.

Dieses Bild hat ein leeres Alt-Attribut. Der Dateiname ist Staubkreise-1024x781.jpg
Figure 1: Actuation of the exhibit

And now the mathematics of it:

But what is the reason for this phenomenon? For this purpose we want to introduce some mathematical terms which allow a precise description of the observed. For this purpose we first introduce the so-called Euclidean distance on the plane \mathbb R^2: Let \mathbf p=(p_1,p_2) and \mathbf q=(q_1,q_2) be two points in the plane. Then their Euclidean distance is defined as d(\mathbf p,\mathbf q)\coloneqq\sqrt{(p_1-q_1)^2+(p_2-q_2)^2}. A mapping f\colon\mathbb R^2\to\mathbb R^2 of the plane in itself preserves the Euclidean distance, i.e., d(f(\mathbf p),f(\mathbf q))=d(\mathbf p,\mathbf q) holds for all \mathbf p,\mathbf q\in\mathbb R^2. Why do we consider such mappings? Well, the reason is that moving the upper plastic disk against the lower one corresponds exactly to such a distance-preserving self-mapping of the plane, because in this case the distance between two points remains the same, no matter whether we move the disk.

The question the exhibit indirectly asks the visitor is this: What distance-preserving self-images of the Euclidean plane are there anyway?

We will fully address this question in the course of this short in-depth text. To do this, we first observe that any translation around a given vector \mathbf v=(v_1,v_2)\in\mathbb R^2 is distance-preserving, because it holds

    \[d(\mathbf p+\mathbf v,\mathbf q+\mathbf v)=\sqrt{(p_1+v_1-(q_1+v_1))^2+(p_2+v_2-(q_2+v_2))^2}=\sqrt{(p_1-q_1)^2+(p_2-q_2)^2}=d(\mathbf p,\mathbf q).\]

But the translations are now in bijection with the vectors \mathbf v\in\mathbb R^2 themselves. So if f\colon\mathbb R^2\to\mathbb R^2 is a distance-preserving mapping, we can form the mapping f'=f-f(\mathbf 0) which still preserves the Euclidean distance but now fixes the zero point \mathbf 0. Therefore, we may assume that f(\mathbf 0)=\mathbf 0. Now let us consider the images of the two standard unit vectors \mathbf e_1 and \mathbf e_2 under f. The right triangle \mathbf 0\mathbf e_1\mathbf e_2 is determined by the lengths of its sides (namely 1, 1 and \sqrt{2}) except for congruence. However, these are preserved under f, so f(\mathbf e_1) and f(\mathbf e_2) must again be unit vectors perpendicular to each other. Consider further that each vector \mathbf v=(v_1,v_2)\in\mathbb R^2 is uniquely determined by its distances to the three points \mathbf 0, \mathbf v_1, \mathbf v_2. Namely, these are exactly a=\sqrt{v_1^2+v_2^2}, b=\sqrt{(v_1-1)^2+v_2^2}, c=\sqrt{v_1^2+(v_2-1)^2}, so that a^2-b^2=2v_1-1 and a^2-c^2=2v_2-1 uniquely fix the parameters v_1 and v_2. But this must give f(\mathbf v)=v_1f(\mathbf e_1)+v_2f(\mathbf e_2), because this vector has the same distances to \mathbf 0=f(\mathbf 0), f(\mathbf e_1), f(\mathbf e_2) as v to \mathbf 0,\mathbf v_1,\mathbf v_2. Thus the mapping f must now even be linear (because this corresponds exactly to the observation just made). But since f also maps the orthonormal basis \mathbf e_1,\mathbf e_2 to such a basis again, f corresponds to an orthogonal 2\times 2 matrix. However, these matrices can now be described very simply: To do this, suppose that \mathbf e_1 maps to the unit vector \mathbf f_1=(a,b). Because a^2+b^2=1, we find an angle \alpha\in[0,2\pi) such that a=\cos(\alpha) and b=\sin(\alpha) (since the equation exactly describes the unit circle). The vector \mathbf e_2 must now be mapped to a unit vector \mathbf f_2=(c,d) which is perpendicular to \mathbf f_1. But it already follows that \mathbf f_2=\pm(-b,a), since there are only two such vectors. Thus \mathbf f_2=(c,d)=(\mp\sin(\alpha),\pm\cos(\alpha)) holds. So for f we get the matrix

    \[M=\begin{pmatrix} \cos(\alpha) & -\varepsilon\sin(\alpha)\ \sin(\alpha) & \varepsilon\cos(\alpha)\end{pmatrix}.\]

Here \varepsilon\in{\pm1} is to be chosen arbitrarily. If now \varepsilon=+1, this simply corresponds to a rotation by the angle \alpha. This is the fact brought out by the exhibit. If, on the other hand, \varepsilon=-1, we obtain a reflection on the straight line g intersecting the x axis at an angle of \alpha/2.

Thus we have fully understood the distance-preserving self-mappings of the Euclidean plane: Any such mapping f\colon\mathbb R^2\to\mathbb R^2 is of the form

    \[f(\mathbf x)=D_{\alpha,\varepsilon}\mathbf x+\mathbf v\]

for a rotation or reflection matrix D_{\alpha,\varepsilon} as above and a vector \mathbf v corresponding to the subtracted translation. However, if we start at the identical self-mapping (where \varepsilon=+1 holds) and take it to a certain other distance-preserving mapping, the parameter \varepsilon cannot make a “jump” (since otherwise the orientation would suddenly reverse and thus the plate would be flipped, which is not allowed in the exhibit). So we can limit ourselves to the case \varepsilon=+1 (if \varepsilon=-1, we can show that then f always represents the reflection at a straight line). Now two cases can occur:

Case 1: f has a fixed point \mathbf x. Then f(\mathbf x)=\mathbf x=D_{\alpha,1}\mathbf x+\mathbf v holds. From this follows (\mathrm{id}-D_{\alpha,1})\mathbf x=\mathbf v. So f is then given by f(\mathbf y)= D_{\alpha,1}\mathbf y+\mathbf x-D_{\alpha,1}\mathbf x=D_{\alpha,1}(\mathbf x-\mathbf y)+\mathbf x, which is simply the rotation around the point \mathbf x by the angle \alpha.

Case 2: f has no fixed point. Since the equation \mathbf x=D_{\alpha,1}\mathbf x+\mathbf v is always solvable for \alpha\neq 0 (because \mathrm{id}-D_{\alpha,1} is then not singular), consequently \alpha=0 must hold. f is then simply a translation.

The two cases 1 and 2 now completely clarify the phenomenon observed at the exhibit: Either one sees large dust circles (case 1) or at least dust straight lines (case 2).

Finally, it should be noted that observed interesting fact for the group of all orientation-preserving (\varepsilon=+1) self-mappings of the Euclidean plane lies group-theoretically in the fact that it is a so-called Frobenius group. This is expressed precisely in the fact that the elements which do not fix a point together with the identity form a subgroup (the translations).


Literature

[1] https://de.wikipedia.org/wiki/Drehgruppe

[2] https://de.wikipedia.org/wiki/Euklidischer_Raum

[3] https://de.wikipedia.org/wiki/Orthonormalbasis

[4] https://de.wikipedia.org/wiki/Frobeniusgruppe