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:

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

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

$z=r \cos\theta$

  • 1
    $\begingroup$ 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. $\endgroup$ 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 $\phi$. $\endgroup$ 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


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.