# Converting rx in spherical coordinates to cartesian.

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.

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quasar987
Homework Helper
Gold Member
what's rx?

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

HallsofIvy
Homework Helper
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.

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.

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]

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).

Now if you want the x-component of $\vec S$ you use the dot product, $\vec S \cdot \hat x$.

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.