Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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


    User Avatar
    Science Advisor

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook