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In cartesian coordinates, a rotation around the z axis is represented as:

$$\left( \begin{array}{ccc} \cos\phi & \sin\phi & 0\\ -\sin\phi & \cos\phi & 0\\ 0&0&1 \end{array} \right)$$ What is the analogue in spherical coordinates?

EDIT My change of coordinates is:

$x=r \sin\theta\cos\phi$

$y=r \sin\theta\sin\phi$

$z=r \cos\theta$

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    $\begingroup$ Perhaps that you just add a constant to the angle $\varphi$? $\endgroup$
    – Matti P.
    Dec 31, 2020 at 10:09
  • $\begingroup$ @MattiP. But in spherical coordinates a vector is represented in the basis $$\left(\begin{array}{c} r\\ \theta\\ \phi \end{array}\right)$$ this confuses me. $\endgroup$
    – mattiav27
    Dec 31, 2020 at 10:15
  • $\begingroup$ I started to write out an answer, but realized that there are too many conventions that are ambiguous here. For example, your rotation is clockwise as viewed looking down the $z$-axis towards the origin, sending positive $+x \mapsto -y$ and $+y \mapsto +x$ for $\phi = +\frac{\pi}{2}$, which is the reverse of the standard convention. $\endgroup$
    – Sammy Black
    Dec 31, 2020 at 10:21
  • $\begingroup$ Please edit your question to include the explicit change-of-coordinates from spherical to cartesian that you're using, i.e. give each of $x$, $y$, and $z$ in terms of $r$, $\theta$, and $\phi$. $\endgroup$
    – Sammy Black
    Dec 31, 2020 at 10:23
  • $\begingroup$ @SammyBlack I have edited the question as required $\endgroup$
    – mattiav27
    Dec 31, 2020 at 10:38

1 Answer 1

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We wish to rotate a vector $\vec{r}$ to $R\vec{r}$ for a suitable rotation matrix $R$. I should make an important conceptual correction: the correct representation of $\vec{r}$ is $r\hat{r}$, not $r\hat{r}+\theta\hat{\theta}+\phi\hat{\phi}$ or whatever you were thinking of. In particular$$\begin{align}\hat{r}&=\sin\theta\cos\phi\hat{i}+\sin\theta\sin\phi\hat{j}+\cos\theta\hat{k},\\\hat{\theta}&=\cos\theta\cos\phi\hat{i}+\cos\theta\sin\phi\hat{j}-\sin\theta\hat{k},\\\hat{\phi}&=-\sin\phi\hat{i}+\cos\phi\hat{j}\end{align}$$is our orthonormal basis. We can write these equations as $\hat{S}_A=T_{Aa}\hat{C}_a$ (I sum over repeated indices), with $\hat{C}_a$ ($\hat{S}_A$) the Cartesian (spherical) basis vectors, e.g. $T_{ri}=\sin\theta\cos\phi$. Now you just want to compute $R_{AB}=T_{Aa}R_{ab}(T^{-1})_{bB}$, or in matrix terms $R^\text{spherical}=TR^\text{Cartesian}T^{-1}$; I'll leave that to you.

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