# Obtaining spherical coordinates by rotations

1. Jul 10, 2013

### Niles

Hi

Say I have a point on a unit sphere, given by the spherical coordinate $(r=1, \theta, \phi)$. Is this point equivalent to the point that one can obtain by $(x,y,z)=(1,0,0)$ around the $y$-axis by an angle $\pi/2-\theta$ and around the $z$-axis by the angle $\phi$?

I'm not sure this is the case, since the spherical coordinate $\phi$ is merely a projection, but I would like to hear your opinion.

2. Jul 10, 2013

### HallsofIvy

Rotation about the y axis through an angle $\pi/2- \theta$ can be calculated by multiplying the position vector of the point by the matrix
$$\begin{bmatrix}cos(\pi/2- \theta) & 0 & -sin(\pi/2- \theta) \\ 0 & 1 & 0 \\ sin(\pi/2- \theta & 0 & cos(\pi/2- \theta)\end{bmatrix}$$
Of course, $$cos(\pi/2- \theta)= sin(\theta)$$ and $$sin(\pi/2- \theta)= cos(\theta)$$ so that is
$$\begin{bmatrix}sin(\theta) & 0 & -cos(\theta) \\ 0 & 1 & 0 \\ cos(\theta) & 0 & sin(\theta) \end{bmatrix}$$

and rotation through angle $\phi$ about the z axis is given by
$$\begin{bmatrix}cos(\phi) & -sin(\phi) & 0 \\ sin(\phi) & cos(\phi) & 0 \\ 0 & 0 & 1\end{bmatrix}$$

So starting from (1, 0, 0) and rotating as you say, you have
$$\begin{bmatrix}cos(\phi) & -sin(\phi) & 0 \\ sin(\phi) & cos(\phi) & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}sin(\theta) & 0 & -cos(\theta) \\ 0 & 1 & 0 \\ cos(\theta) & 0 & sin(\theta) \end{bmatrix}\begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix}$$
$$\begin{bmatrix}cos(\phi) & - sin(\phi) & 0 \\ sin(\phi) & cos(\phi) & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}sin(\theta) \\ 0 \\ cos(\theta) \end{bmatrix}$$

Of course, in polar coordinates, $$x= cos(\phi)sin(\theta)$$, $$y= sin(\phi)sin(\theta)$$, $$z= cos(\theta)$$ is just the usual coordinates with x and y reversed. That is, your two rotations swap the x and y axes.

3. Jul 10, 2013

### Mandelbroth

The coordinate transformation from spherical coordinates to Cartesian coordinates is given by $(r,\theta,\varphi)\rightarrow(r\sin\theta\cos\varphi,r\sin\theta\sin \varphi,r\cos\theta)$.

4. Jul 11, 2013

### Niles

Thanks for your help, yeah, I guess I should just have rotated the coordinate myself to see that indeed that is how a spherical point is obtained. Thanks again.