Distance Formula in Curvilinear Coordinates

• This problem is used in the following sequences

  1. << Magnetic Field Due to a Spinning Ring of Charge | E&M Ring Cycle Sequence |

• Static Fields 2021

  The distance \(\left\vert\vec r -\vec r\,{}'\right\vert\) between the point \(\vec r\) and the point \(\vec r'\) is a coordinate-independent, physical and geometric quantity. But, in practice, you will need to know how to express this quantity in different coordinate systems.

  Hint: Be sure to use the textbook: https://books.physics.oregonstate.edu/GSF/coords2.html

  1. Find the distance \(\left\vert\vec r -\vec r\,{}'\right\vert\) between the point \(\vec r\) and the point \(\vec r'\) in rectangular coordinates.

     

  2. The distance \(\left\vert\vec r -\vec r\,{}'\right\vert\) between the point \(\vec r\,{}'=(x\,{}',y\,{}',z\,{}')\) and the point \(\vec r=(x,y,z)\) is a coordinate-independent, physical and geometric quantity. But, in practice, you will need to know how to express this quantity in different coordinate systems. Show that this same distance written in cylindrical coordinates is: \begin{equation*} \left|\vec r -\vec r\,{}'\right| =\sqrt{s^2+s\,{}'^2-2ss\,{}'\cos(\phi-\phi\,{}') +(z-z\,{}')^2} \end{equation*}
  3. Show that this same distance written in spherical coordinates is: \begin{equation*} \left\vert\vec r -\vec r\,{}'\right\vert =\sqrt{r'^2+r\,{}^2-2rr\,{}' \left[\sin\theta\sin\theta\,{}'\cos(\phi-\phi\,{}') +\cos\theta\cos\theta\,{}'\right]} \end{equation*}
  4. Now assume that \(\vec r\,{}'\) and \(\vec r\) are in the \(x\)-\(y\) plane. Simplify the previous two formulas.