# How can the trefoil knot be expressed in polar coordinates?

From Wikipedia, the parametric equations for a trefoil knot are

\begin{align*} x(t) &= \sin t + 2\sin 2t \\ y(t) &= \cos t - 2\cos 2t \\ z(t) &= -\sin 3t. \end{align*}

I am only interested in the $x$ and $y$ dimensions, so $z(t)$ is ignored. When I plot it with Wolfram|Alpha, I get the expected general shape. However, when I try to convert it to polar coordinates, it (seemingly) just doesn't work.

\begin{align*} r^2 &= x^2 + y^2 \\ &= (\sin t + 2\sin 2t)^2 + (\cos t + 2\cos 2t)^2 \\ &= (\sin^2 t + 4\sin t \sin 2t + 4\sin^2 2t) + (\cos^2 t - 4\cos t \cos 2t + 4\cos^2 2t) \\ &= 1+4 + 4(\sin t \sin 2t - \cos t \cos 2t) \\ &= 5-4\cos 3t \end{align*}

Yet, when I try to plot $r = \sqrt{5-4\cos 3t}$, I get something completely different. What's the problem? Additionally, how could you express the trefoil knot in polar coordinates?

• The problem is that there is no reason why you should have $t = \theta$. When using polar coordinates you plot the radius in terms of the angle. So $r^2 = x(\theta)^2 + y(\theta)^2$. But you're using $r^2 = x(t)^2 + y(t)^2$ where $t$ is just some probably arbitrary parametrization parameter of the trefoil. – user38355 Nov 22 '13 at 18:35
• But the Wolfram|Alpha plots both treat $t$ analogously to $\theta$. In the first, $0 < t < 2\pi$ and in the second $-\pi < t < \pi$. How is $t$ different from $\theta$ in these contexts? – El'endia Starman Nov 22 '13 at 18:47

When you do polar plots you are stuck with parametrizations of the limited form $$x(\varphi)=r(\varphi)\cos \varphi, \quad y(\varphi)=r(\varphi)\sin(\varphi).$$ The parametrization that you gave is not of that form. This is apparent already from the observation that the parametrization gives the full trefoil, when $t\in[0,2\pi]$, but the trefoil wraps around the origin twice, so $\varphi$ should range over $[0,4\pi]$.

Let us look at the Wikipedia parametrization of the trefoil on the surface of a torus: $$x=(2+\cos3t)\cos2t,\quad y=(2+\cos3t)\sin2t,\quad z=\sin 3t.$$ If we ignore that $z$-coordinate for a moment, we see $(x,y)\uparrow\uparrow(\cos 2t,\sin 2t)$, which is a tell-tale sign that here $\varphi=2t$ is the polar angle coordinate. As $t$ ranges over $[0,2\pi]$, we should, indeed, have $\varphi\in[0,4\pi]$. Thus the projection of that trefoil onto the $xy$-plane comes from the polar equation $$r=2+\cos\frac{3\varphi}2,$$ as suggested by heropup (+1). The plot is not quite what you may have expected:

Here the additive constant $2$ represents the ratio of the radius of the "wire" inside the torus to that of the "tube" around the wire. IMVHO the projection looks a bit cleaner, if we use ratio $4$ and equation $r=4+\cos\frac{3\varphi}2$ instead:

For a better view here is a 3D-image of how the trefoil wraps itself around the torus.

The trefoil is the thin tube on the surface of the doughnut.

The first comment is correct.

Your $t$ is not the same as $\theta$ for a given point on the trefoil knot. That $\theta$ is $\arctan(\frac{y(t)}{x(t)})$. If you look at the graph of a trefoil knot you can see there can't be any polar equation for it because the mapping from $\theta$ to $r$ is not one-to-one. The best you could do is a parametric equation in polar coords, which would just be

$$\theta(t) = \arctan\frac{\cos(t)-2\cos(2t)}{\sin(t) + 2\sin(2t)}\\ r(t) = \sqrt{5-4\cos t(3t)}$$

And I'm not even sure that will give you the full trefoil knot- it could depend on which branch of $\arctan$ is chosen.

You could try to solve for $t$ in terms of $\theta$ (looks hard), then plug that into the formula for $r$, but at best you'll only get one correct $r$ value for each $\theta$, so it won't be the full trefoil knot.

• It's not quite right to say there can't be a polar equation, since as in heropup's example theta could go beyond 360 degrees and wrap around giving new values of r. So maybe there is a polar equation for it after all. But it looks complicated to reproduce this exact parametric representation in polar coords. – Toby May 21 '14 at 17:55

An epitrochoid in two dimensions can yield a projection of a trefoil knot, or you an just make an educated guess: e.g., $$r = 2 + \sin \tfrac{3}{2}\theta, \quad \theta \in [0, 4\pi).$$

This question is of interest to me, too. I do not have an answer to the question above, but can provide some additional information. First, if one enters sqrt((sint+2sin2t)2+(cost+2cos2t)2) into Wolfram Alpha, it returns a different derivation, which when plotted does not look anything like sqrt(5−4cos3t). Second, I think the problem with the logic above is that some information must be lost in the process of squaring the statements. By playing around in gnuplot, the closest I've been able to get to a knot is 1+0.05cos(3t+0.5t) over -2pi..2pi, 1+0.05cos(3t+0.3t) over -3pi..3pi. One can play with the scaling of t as desired (primes just as in knot theory?) but the range has to be adjusted accordingly. I hope someone else can put this in the context of the original question. Regards, David