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I have a question about converting functions defined in Cartesian coordinates to a cylindrical polar system. The particular coordinate transformation that I'm reading about is:

\begin{equation} \begin{pmatrix}x\\[0.3em]y\\[0.3em]z\end{pmatrix} = - \begin{pmatrix}\pi/4\\[0.3em] \pi/4\\[0.3em] \pi/4\end{pmatrix} + \begin{pmatrix} 1/\sqrt{2}\quad & 1/\sqrt{6}\quad & 1/\sqrt{3}\,\, \\[0.3em] -1/\sqrt{2}\quad & 1/\sqrt{6} \quad & 1/\sqrt{3}\,\, \\[0.3em] 0\quad & -2/\sqrt{6}\quad & 1/\sqrt{3} \,\, \end{pmatrix} \begin{pmatrix}\rho\cos\theta\\[0.3em] \rho\sin\theta \\[0.3em] \zeta \end{pmatrix}, \end{equation}

which I assume can be rewritten as

\begin{equation} \begin{pmatrix}\rho\cos\theta\\[0.3em] \rho\sin\theta \\[0.3em] \zeta \end{pmatrix} = \begin{pmatrix} 1/\sqrt{2}\quad & -1/\sqrt{2}\quad & 0\,\, \\[0.3em] 1/\sqrt{6}\quad & 1/\sqrt{6} \quad & -2/\sqrt{6}\,\, \\[0.3em] 1/\sqrt{3}\quad & 1/\sqrt{3}\quad & 1/\sqrt{3} \,\, \end{pmatrix} \begin{pmatrix}x+\pi/4\\[0.3em]y+\pi/4\\[0.3em]z+\pi/4\end{pmatrix} \end{equation}

The paper I'm reading states that $f(x,y,z) = (z,x,y)$ and so maps $(\rho,\theta,\zeta)$ to $(\rho,\theta+2\pi/3,\zeta)$ and I can see this is the case, as $f$ simply rotates around the axis $x=y=z$ (which is around the $\zeta$ axis, I think). However, I'm unsure of quite how to show this analytically.

There is another function $g(x,y,z) = (-y-\pi/2,-x-\pi/2,-z-\pi/2)$, which according to the paper, maps $(\rho,\theta,\zeta)$ to $(\rho,-\theta,-\zeta)$ but this one is far less obvious to me.

I assume this is a simple procedure to convert the function to the new coordinate system and that I'm missing the trick! Any ideas?

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  • $\begingroup$ Your inverse matrix should have $1/\sqrt6$ as its central entry, I think. The $\eta$ instead of $\zeta$ might be a typo as well. It would help to know what paper you are reading. $\endgroup$ – MvG Mar 6 '14 at 14:55
  • $\begingroup$ You're very right - I mistyped those both and have edited them. It's a paper regarding magnetic fields in a periodic box driven by a highly symmetric flow. The two functions are symmetries of the flow but there are more. I found the paper at: tandfonline.com/doi/abs/10.1080/03091929.2013.832762 $\endgroup$ – tiit_helimut Mar 6 '14 at 18:24
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Function f

I'd add a $1$ as a fourth coordinate (homogenous coordinates) so that you can include the addition in your matrix multiplication. Then you'd have

$$M = \begin{pmatrix} 1/\sqrt2 & 1/\sqrt6 & 1/\sqrt3 & -\pi/4 \\ -1/\sqrt2 & 1/\sqrt6 & 1/\sqrt3 & -\pi/4 \\ 0 & -2/\sqrt6 & 1/\sqrt3 & -\pi/4 \\ 0 & 0 & 0 & 1 \end{pmatrix} \qquad M^{-1}=\begin{pmatrix} 1/\sqrt2 & -1/\sqrt2 & 0 & 0 \\ 1/\sqrt6 & 1/\sqrt6 & -2/\sqrt6 & 0 \\ 1/\sqrt3 & 1/\sqrt3 & 1/\sqrt3 & \pi\sqrt3/4 \\ 0 & 0 & 0 & 1 \end{pmatrix} $$

Now you can express $f$ like this:

$$ \begin{pmatrix} \rho'\cos\theta' \\ \rho'\sin\theta' \\ \zeta' \\ 1 \end{pmatrix} = M^{-1}\begin{pmatrix} 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}M \begin{pmatrix} \rho\cos\theta \\ \rho\sin\theta \\ \zeta \\ 1 \end{pmatrix} = \begin{pmatrix} -1/2 & -\sqrt3/2 & 0 & 0 \\ \sqrt3/2 & -1/2 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \rho\cos\theta \\ \rho\sin\theta \\ \zeta \\ 1 \end{pmatrix} $$

So the identity $\zeta'=\zeta$ is very obvious. In the top left $2\times2$ part of the combined matrix you see this rotation you are talking about. You see that the matrix is orthogonal, so it will preserve lenths, therefore $\rho' = \rho$. You can also compute the angle of rotation $\varphi$ using

\begin{gather*} \cos\varphi = -\frac12 \qquad \sin\varphi = \frac{\sqrt3}2 \\ \varphi = \arctan\frac{\frac{\sqrt3}2}{-\frac12}+\pi = \arctan(-\sqrt3)+\pi = -\frac13\pi+\pi = \frac23\pi \end{gather*}

As usual when computing an angle from values for $\sin$ and $\cos$, you have to take care of the fact that $\arctan$ is only defined up to multiples of $\pi$. In computer programs there is often a function atan2 to take care of this fact.

Function g

Now the same for $g$:

$$ \begin{pmatrix} \rho''\cos\theta'' \\ \rho''\sin\theta'' \\ \zeta'' \\ 1 \end{pmatrix} = M^{-1}\begin{pmatrix} 0 & -1 & 0 & -\pi/2 \\ -1 & 0 & 0 & -\pi/2 \\ 0 & 0 & -1 & -\pi/2 \\ 0 & 0 & 0 & 1 \end{pmatrix}M \begin{pmatrix} \rho\cos\theta \\ \rho\sin\theta \\ \zeta \\ 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \rho\cos\theta \\ \rho\sin\theta \\ \zeta \\ 1 \end{pmatrix} $$

The sign change $\zeta''=-\zeta$ is easy to see. You can also see that the length $\rho''=\rho$ is preserved, but the angle $\theta''=-\theta$ changes its sign. If you want to do this formally as well, you'd again spot the orthogonal matrix, and identify it as a reflection not a rotation.

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  • $\begingroup$ Thank you for taking the time to explain this to me - your solution is clear and well written! As this is not my field of study, forgive me if I'm incorrect, but you are transforming back to the original (cartesian) coordinate system, carrying out the operation (either $f$ or $g$) and then transforming to the polar coordinate system. I hadn't thought to include the translations in such a way! I would upvote your answer but I do not have enough reputation. Thank you again! $\endgroup$ – tiit_helimut Mar 6 '14 at 18:30
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    $\begingroup$ Well, even though the parameters $\rho,\theta,\zeta$ are polar (i.e. cylindric), the vector we both used is not. So I'd view both the $(\rho\cos\theta,\rho\sin\theta,\zeta)$ and $(x,y,z)$ vectors as Cartesian. Which makes sense, since the transformation between polar and Cartesian is not linear. The step back to a conclusion about the polar parameters requires some interpretation of the resulting matrices, which is what all this stuff about orthogonality and angle of rotation and so on is about. But having everything else neatly taken care of makes that step reasonably clear as well. $\endgroup$ – MvG Mar 6 '14 at 18:39

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