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I hope that I can explain myself clear enough, Assuming i have a sphere that has been moved down in the Z-axis. I know that the radial unit vector when the sphere is not shifted can be expressed as: $$\hat{r} = \sin\theta\cos\phi \,\hat{x} + \sin\theta\sin\phi \,\hat{y} + \cos\theta\,\hat{z}$$ where $\hat{x}$, $\hat{y}$, and $\hat{z}$ are the Cartesian unit vectors. But how can I obtain $\hat{r}$ over the surface of the sphere in a situation when the sphere has been shifted? like in the picture I added?enter image description here

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    $\begingroup$ Might Mathematics be better suited for this math question? $\endgroup$ – Kyle Kanos Jul 26 '15 at 15:53
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    $\begingroup$ Consider the transformation that took the system from sphere-at-origin to sphere-not-at-origin (i.e. $z' \leftarrow z-2R$). Now consider the reverse transformation... $\endgroup$ – dmckee Jul 26 '15 at 17:05
  • $\begingroup$ I agree with @KyleKanos However short answer: If you have a parametrisation $r(x1,x2,x3)$ you can always find a local tripod via computing $\partial_{x_i} r$ $\endgroup$ – Bort Jul 27 '15 at 9:45
  • $\begingroup$ The way you've drawn $\hat r$ in the diagram, it hasn't changed due to the shift. If you want something else than what you've already written, it needs to be reflected in the diagram. $\endgroup$ – joriki Jul 28 '15 at 9:17

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