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Converting rx in spherical coordinates to cartesian.

  • Thread starter seang
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  • #1
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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.
 

Answers and Replies

  • #2
quasar987
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what's rx?
 
  • #3
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Is is a component of the vector I am working to convert to cartesian coordinates.
 
  • #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.
 
  • #5
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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.
 
  • #6
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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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