5
$\begingroup$

Suppose there are two observational frames of reference with origins $O$ and $O'$, respectively, separated by some constant distance. A body located at point $P$ has Cartesian coordinates $\left(x,y,z\right)$ and $\left(x',y',z',\right)$ in $O$ and $O'$, respectively; and similarly, the spherical coordinates of $P$ in the unprimed and primed frames are $\left(r,\theta,\phi\right)$ and $\left(r',\theta',\phi'\right)$.

Say $O$ determines the Cartesian coordinates of $O'$ to be $\left(a,b,c\right)$. Then the Cartesian coordinates of $P$ in $O$ are simply related to the primed Cartesian coordinates: $$\left(x,y,z\right)=\left(x'+a,y'+b,z'+c\right)$$

My question, then, is how does one compute the affect of translation on spherical coordinates. That is, given $\left(a,b,c\right)$, how can one write $\left(r,\theta,\phi\right)$ as a function of $\left(r',\theta',\phi'\right)?$

$\endgroup$
3
$\begingroup$

There is kind of an answer over at Math. All you can do in spherical coordinates is to change the position of your "pole", i.e. you have $(1,0,0)$ in your first coordinate system, which is mapped to some $(r', \vartheta', \varphi')$ in the second coordinate system. The two angles represent a rotation, and the $r$ represents a scaling.

I think since we still deal with a vector space (for the angular part), you can simply add the origin of coordinate system two to vectors of the first coordinate system. The radius is a scaling, and needs to be multiplied. Hence for a point $p=(r, \vartheta, \varphi)$ in coordinate system 1 you get:

$$ p'=(r \cdot r', \vartheta+\vartheta', \varphi+\varphi') $$

$\endgroup$
  • $\begingroup$ Reading this, I am not quite sure if maybe the $r'$ needs to be multiplied instead, since it represents a scaling... $\endgroup$ – Arne Dec 10 '13 at 10:04
  • $\begingroup$ I am rather sure now. Please correct me if I'm wrong -- edited the answer accordingly. $\endgroup$ – Arne Dec 10 '13 at 10:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.