# On the curve generated by plotting one sine against another

A circle? A line? Actually, it depends. It depends upon parameters like the frequency or the phase; when these change, really interesting things happen. What I mean by “against another” is that the first sine function will be the $x$-coordinate while the other, the $y$-coordinate; then, if you studied engineering you might say “$y$ vs. $x$” but if you studied biology you might be used to say “$x$ vs. $y$” (personal observation). We end up then with a curve, say $\gamma(t) = (x(t), y(t))$, where

$x(t) = A_1\sin(\omega_1 t + \phi_1)$

and

$y(t) = A_2\sin(\omega_2 t + \phi_2)$.

Remember $\omega_i$ is the frequency (how many wiggles in a given interval), $\phi_i$ is the phase (horizontal displacement) and $A_i$ is the amplitude ($i \in \{1,2\}$). Starting with the simplest case, let’s slide or shift one sine wave while keeping the other still ($\phi_1 = 0$); both functions will have the same frequency($\omega_1 = \omega_2$) but one changes it’s phase. As you can see in the following animation, the corresponding curve changes from a line to a circle, passing through an ellipse, and back. In the figure, the individual coordinates are on the left and the corresponding curve is in the right. Superimposed on the curve there is a trace of the previous curves just for fun.

How come? Is it obvious that this should be the case? Is there another way of understanding this? Maybe. If you squint a little bit, you will see that the changes in shape in the ellipse resemble some funny rotation, as if there was an hidden dimension we are not seeing. It looks like if there were some sort of cylinder the shape was confined to. Let’s try that: I am going to add another dimension and see whether the kind of rotation I am imagining explains the changes in shape.

The starting point is an ellipse in the $xy$ plane. Let’s turn it a little by an angle $\alpha$ in the new $z$ direction and start rotating it around the circle with radius $r$. Mathematically, this means that, the ellipse whose parametric equations are

$x = A_1 \cos\theta$

$y = A_2 \sin\theta$

$z = 0$,

is to be multiplied by the rotation matrices:

$\mathbf{R} =\begin{pmatrix}1 & 0 & 0 \\0 & \cos\alpha & -\sin\alpha\\0 & \sin\alpha & \cos\alpha \end{pmatrix}$

and

$\mathbf{Q} =\begin{pmatrix}\cos\phi & 0 & -sin\phi \\0 & 1 &0\\\sin\phi & 0 & \cos\phi \end{pmatrix}$,

in that order. The result is:

$x_{rq} = \frac{A_1 - A_2\sin\alpha}{2}\cos(t - \phi) + \frac{A_1 + A_2\sin\alpha}{2}\cos(t + \phi)$

$y_{rq} = A_2\sin t \cos\alpha$,

We don’t care about the $z_{rq}$ coordinate. Therefore, if $A_1 = A_2\sin\alpha$, the rotation in this extended euclidean space IS indeed a change of phase in the two dimensional projection! The parameter $\alpha$ is free so we can always find the corresponding higher dimensional ellipse given one in a hyperplane. The following animation illustrates what I just did. To the left you see the rotating $3$D ellipse and its projection. Four points have been marked in order to keep track of its trajectories. To the right, the individual sine waves illustrating the phase change.

So far so good, however, I was wrong in one detail: the ellipse is not confined to the cylinder, it seems to be wrapped around the cylinder! That observation was first made, apparently, by one of the important figures of dynamical systems theory, Vladimir Arnold[1], who used it to explain a very peculiar family of curves: the Lissajous Curves. There is a recent blog post that has some nice animations[2], but unfortunately it is in Turkish and I don’t speak Turkish, but, if you do, go and read it! Let me try to explain what this is all about.

We are still working with the two sine waves introduced at the beginning. But now, we are going to allow one of the frequencies, say $\omega_2$, to vary as well as the phase $\phi_2$; $\omega_1 =1$. Take a band of height $2A_2$, draw a sine wave with amplitude $A_2$ and period $\frac{2\pi A_1}{\omega_2}$, that is

$s(t) = A_2 \sin \frac{\omega_2}{A_1}t$,

and wrap it around a cylinder whose base has radius $A_1$. In the case of $\omega_2 = 1$, the result is something like this:

The “rolling” or winding is achieved by rolling the time domain around a circle of radius $A_1$ and let the height be the actual function. Mathematically that is:

$(t, s(t)) \mapsto (A_1\cos t, A_1 \sin t, s(t))$.

It shouldn’t be a surprise that, in this case, an ellipse appears in the cylinder; after all, the period is the same as the perimeter of the circle ($2\pi\mbox{radius}$).

As a final step, choose any hyperplane that contains the axis of the cylinder and project the resulting curve on it. In this case I chose a plane parallel to the $yz$ plane. If you rotate the cylinder, you get the same change of shape we observed before.

The pleasant surprise comes when you change $\omega_2$, that is, instead of sliding two sine wave of the same frequency, you allow them to differ. In that case we get a variety of curves that depend upon the frequency difference. For example, for  $\omega_2 \approx 1$, the figure looks like going along all the ellipses obtained before but never settling in one (Why is this? can you think of a mathematical reason for that?)

For smaller values $\omega_2 = 0.7$ (left) and $\omega_2 = 0.5$ (right) below, the trajectory deforms into a closed algebraic curve very similar to the Viviani’s curve, is it the Viviani’s curve? I will talk about that in a next post.

Indeed, according to [1], for every rational number $p = m/n$, the corresponding trajectory is an algebraic closed curve. When $\omega_2$ is irrational, the curve is a space filling curve! We shall prove this in a future post. As an example, I put the corresponding trajectories for $\omega_2 = 5/4$ (left) and $\omega_2 = \pi$.

Finally, the case $\omega_2 = 2$ is worth mentioning because its projection is a parabola but also because it is obtained as the intersection of the cylinder with a parabolic hyperboloid.

Anyways, for now I got tired of writing but in the second version of this post I will show the mentioned theorems and talk about interesting relationships of these curves with other problems. It is worth to mention that you can make these constructions at home! Just check the following paper [3].

References

[1] Arnol’d, V. I. (2013). Mathematical methods of classical mechanics (Vol. 60). Springer Science & Business Media.

[2] Bilim ne guzellan. (2017). Temel Harmonikler ve Lissajous Eğrileri  [Blog post, turkish]. Retrieved from http://bilimneguzellan.net/index.php/2017/12/17/temel-harmonikler-ve-lissajous-egrileri/

[3] Criado, C., & Alamo, N. (2004). A Simple Construction to Illustrate Lissajous Figures. The Physics Teacher, 42(4), 248-249.