Recentering a Spherical Coordinate Sytem How do you recenter a spherical coordinate system. For example, if the center were at $\left (0, 0, 0 \right )$ and I wanted to move the center of the spherical coordinate system to $\left (\rho_{1}, \Theta_{1}, \Phi_{1} \right )$, then what transformation would I apply to $\left (\rho_{2}, \Theta_{2}, \Phi_{2} \right )$?
In cartesian coordinates, you would simply subtract the two vectors.
 A: This is going to be unsightly...
The following Mathematica code:

Needs["VectorAnalysis`"]
Simplify@ CoordinatesFromCartesian[
 CoordinatesToCartesian[{r, theta, phi}, Spherical] 
                     + CoordinatesToCartesian[{r0, theta0, phi0}, Spherical],
 Spherical
 ] 

gives the following output (doctored so that it looks nicer):
$$ r' = \sqrt{r^2+2 r_0 r \left(\sin (\theta ) \sin
   \left(\theta _0\right) \cos \left(\phi -\phi
   _0\right)+\cos (\theta ) \cos \left(\theta
   _0\right)\right)+r_0^2} $$
$$ \theta' = \cos ^{-1}\left(\frac{r \cos (\theta )+r_0 \cos
   \left(\theta _0\right)}{\sqrt{r^2+2 r_0 r
   \left(\sin (\theta ) \sin \left(\theta
   _0\right) \cos \left(\phi -\phi
   _0\right)+\cos (\theta ) \cos \left(\theta
   _0\right)\right)+r_0^2}}\right) $$
$$ \phi' = \tan ^{-1}\left(r \sin (\theta ) \cos (\phi
   )+r_0 \sin \left(\theta _0\right) \cos
   \left(\phi _0\right),r \sin (\theta ) \sin
   (\phi )+r_0 \sin \left(\theta _0\right) \sin
   \left(\phi _0\right)\right) $$
In this last line, there is a two-argument variant of arctan, which is explained here, for example.
