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Obtaining spherical coordinates by rotations

  1. Jul 10, 2013 #1
    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. jcsd
  3. Jul 10, 2013 #2

    HallsofIvy

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

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

    So starting from (1, 0, 0) and rotating as you say, you have
    [tex]\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}[/tex]
    [tex]\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}[/tex]

    Of course, in polar coordinates, [tex]x= cos(\phi)sin(\theta)[/tex], [tex]y= sin(\phi)sin(\theta)[/tex], [tex]z= cos(\theta)[/tex] is just the usual coordinates with x and y reversed. That is, your two rotations swap the x and y axes.
     
  4. Jul 10, 2013 #3
    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)##.
     
  5. Jul 11, 2013 #4
    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.
     
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