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Converting rx in spherical coordinates to cartesian.

  1. Sep 10, 2006 #1
    I have no idea how to do this. I've tried alot of things but I can never reduce it to solely cartesian coordinates. Is there any hard fast procedure to conversions like this? thanks.
     
  2. jcsd
  3. Sep 10, 2006 #2

    quasar987

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    what's rx?
     
  4. Sep 10, 2006 #3
    Is is a component of the vector I am working to convert to cartesian coordinates.
     
  5. Sep 11, 2006 #4

    HallsofIvy

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    Do you know the formulas for conversion from Cartesian to spherical coordinates?

    [tex]x= \rho cos(\theta)sin(\phi)[/tex]
    [tex]y= \rho sin(\theta)sin(\phi)[/tex]
    [tex]z= \rho cos(\phi)[/tex]

    [tex]\rho= \sqrt{x^2+ y^2+ z^2}[/tex]
    [tex]\theta= arctan(\frac{y}{z})[/tex]
    [tex]\phi= arctan(\frac{z}{\sqrt{x^2+y^2}})[/tex]

    The x-component of a vector is just the x coordinate of the corresponding point.
     
  6. Sep 11, 2006 #5
    Yes I have these. Except where ever yours have rho, I have r. That's ok. So can I just say that (rx), or (px) with your equations, is equal to [tex]x\sqrt{x^2+ y^2+ z^2}[/tex]

    ? That seems too easy.
     
  7. Sep 11, 2006 #6
    Let's say your vector in spherical coordinates is:
    [tex] \vec S = (S_R, S_\theta, S_\phi) [/tex]

    and cartesian,
    [tex] \vec C = (C_x, C_y, C_z) [/itex]

    Now if you want the x-component of [itex] \vec C [/itex] you use the dot product, [itex] \vec C \cdot \hat x [/itex], where [itex] \hat x = (1,0,0) [/itex] (in cartesian coordinates).

    Now if you want the x-component of [itex] \vec S [/itex] you use the dot product, [itex] \vec S \cdot \hat x [/itex].

    You need to express the unit vectors in the different coordinate system though. You can do this with geometry.

    That makes it a little bit more difficult for you.
     
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