Obtaining spherical coordinates by rotations

[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.

 

[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.

 

[Mandelbroth]

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.

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

 

[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.

 

[CellCoree]

Hi there,

Thank you for your question. Obtaining spherical coordinates by rotations can be a bit tricky, but let me try to explain it to you.

First, let's define what spherical coordinates are. They are a system of coordinates used to locate points on a sphere using two angles - $\theta$ and $\phi$ - and a distance, $r$, from the origin. The angle $\theta$ measures the angle between the positive $z$-axis and the point, while $\phi$ measures the angle between the positive $x$-axis and the projection of the point onto the $xy$-plane.

Now, to answer your question, let's look at the rotation around the $y$-axis first. This rotation does indeed change the value of $\theta$, as it rotates the point around the $y$-axis. However, it also changes the value of $\phi$, as it changes the projection of the point onto the $xy$-plane. So, after this rotation, the point will have new values for $\theta$ and $\phi$.

Next, let's look at the rotation around the $z$-axis. This rotation will only change the value of $\phi$, as it rotates the point around the $z$-axis. So, after this rotation, the point will have a new value for $\phi$, but the value of $\theta$ will remain the same.

Therefore, the point obtained by rotating $(1,0,0)$ by $\pi/2-\theta$ around the $y$-axis and by $\phi$ around the $z$-axis will have different values of $\theta$ and $\phi$ compared to the point given in spherical coordinates.

I hope this helps clarify things for you. Let me know if you have any further questions. Keep exploring and learning!