Converting rx in spherical coordinates to cartesian.

[seang]

I have no idea how to do this. I've tried a lot of things but I can never reduce it to solely cartesian coordinates. Is there any hard fast procedure to conversions like this? thanks.

 

[quasar987]

what's rx?

 

[seang]

Is is a component of the vector I am working to convert to cartesian coordinates.

 

[HallsofIvy]

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

$$x= \rho cos(\theta)sin(\phi)$$
$$y= \rho sin(\theta)sin(\phi)$$
$$z= \rho cos(\phi)$$

$$\rho= \sqrt{x^2+ y^2+ z^2}$$
$$\theta= arctan(\frac{y}{z})$$
$$\phi= arctan(\frac{z}{\sqrt{x^2+y^2}})$$

The x-component of a vector is just the x coordinate of the corresponding point.

 

[seang]

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 $$x\sqrt{x^2+ y^2+ z^2}$$

? That seems too easy.

 

[FrogPad]

Let's say your vector in spherical coordinates is:
$$\vec S = (S_R, S_\theta, S_\phi)$$

and cartesian,
[tex]\vec C = (C_x, C_y, C_z) [/itex]<br /> <br /> Now if you want the x-component of $\vec C$ you use the dot product, $\vec C \cdot \hat x$, where $\hat x = (1,0,0)$ (in cartesian coordinates).<br /> <br /> Now if you want the x-component of $\vec S$ you use the dot product, $\vec S \cdot \hat x$.<br /> <br /> You need to express the unit vectors in the different coordinate system though. You can do this with geometry.<br /> <br /> That makes it a little bit more difficult for you.[/tex]