Is there a way of subtracting two vectors in spherical coordinate system without first having to convert them to cartesian or other forms?

Since I have already searched and found the difference between Two Vectors in Spherical Coordinates as,


which I believe the radius of displaced vector. I still didn't get any way to find the theta (angle from positive z axis)and psi(angle from positive x axis). Please help.


The given expression is the distance between two points whose spherical coordinates are given. It is derived on Cartesian coordinate basis considering differences of $r,\theta,\phi$ which is the most convenient method.

If $\rho,\rho^{'},\phi, \phi^{'}, \theta $ are given two values of $ \theta ^{'}$ can be calculated.

  • $\begingroup$ Thank You for your information. But here I need to find the $r,\theta,\phi$ of displaced point. Currently I have $\rho,\rho^{'},\phi, \phi^{'}, \theta, \theta' $ using these is it possible to find the new $r,\theta,\phi$. I also know we cannot simply subtract the final from the original. $\endgroup$ – Aswin Jagadeesh A Nov 9 '18 at 5:22
  • $\begingroup$ If distance $R,\phi, \theta ,\rho$ are given then the center and radius of a sphere locus of the distance $R$ from that given point are fixed, Next if either of $ \theta $ or $\phi$ are given then two values of $\phi. \theta $ can be respectively determined. $\endgroup$ – Narasimham Nov 9 '18 at 9:14
  • $\begingroup$ Apologies, I didn't fully understand what you have said. My motive is to find the displacement vector, that means right now i have two points one original and the other final, which are of course in spherical coordinate. Now I want the vector difference so that I hope I'll get the $\delta r,\delta \phi,\delta \theta $. Don't consider that the two points are in the same sphere. $\endgroup$ – Aswin Jagadeesh A Nov 9 '18 at 9:47
  • $\begingroup$ For more information please refer this link. math.stackexchange.com/a/1365667/613522 This helped me with the cylindrical coordinate case. I Just want a similar solution. $\endgroup$ – Aswin Jagadeesh A Nov 9 '18 at 9:52

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