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Section 1.14 The Position Vector in Curvilinear Coordinates

The position vector field is easier to write algebraically in rectangular coordinates than it is to think about:

\begin{equation} \rr=x\,\xhat+y\,\yhat+z\,\zhat .\label{Position}\tag{1.14.1} \end{equation}

What happens in curvilinear coordinates? Naive pattern matching with (1.14.1) might lead you to believe that the position vector in spherical coordinates is given by:

\begin{equation} \rr=r\,\rhat+\theta\,\that+\phi\,\phat \qquad\qquad\hbox{(incorrect)} .\tag{1.14.2} \end{equation}

However, if you try to follow this equation as a literal set of instructions, then the instructions say “first travel the distance \(r\) in the \(\rhat\) direction (which already gets you to where you want to go) and then from there, travel the distance \(\theta\) (which is an angle, not a distance) in the \(\that\) direction and then from there, travel the distance \(\phi\) (which is also an angle, not a distance) in the \(\phat\) direction.” These directions are clearly incorrect; we should have stopped after the first step. In spherical coordinates, the position vector is given by:

\begin{equation} \rr=r\,\rhat\qquad\qquad\hbox{(correct)} .\tag{1.14.3} \end{equation}

Don't forget that the position vector is a vector field, which depends on the point \(P\) at which you are looking. However, if you try to write the position vector \(\rr(P)\) for a particular point \(P\) in spherical coordinates, and you think of the tail of the position vector as “attached” to the origin, then you have a problem. It is not clear which \(\rhat\text{,}\) \(\that\text{,}\) and \(\phat\) you should use. The resolution to this dilemma, of course, is to think of the tail as “attached” to the point \(P\) and to use the \(\rhat\text{,}\) \(\that\text{,}\) and \(\phat\) at \(P\text{.}\) We can use a subscript \(P\) to denote this particular choice of basis vectors \(\rhat_P\text{,}\) \(\that_P\text{,}\) and \(\phat_P\text{.}\) If you are trying to use the position vector to tell someone how to get to \(P\) from the origin and you try to use curvilinear coordinate basis vectors, it turns out to be somewhat tricky. As discussed above, the process boils down to facing in the correct direction (\(\rhat_P\)), then going the correct distance (\(r\)), so that

\begin{equation} \rr(P)=r\,\rhat_P .\tag{1.14.4} \end{equation}

You should try to use a similar process to find the position vector in cylindrical coordinates. If you are trying to use the position vector to tell someone how to get to \(P\) from the origin and you try to use curvilinear coordinate basis vectors, then they have to know where \(P\) is in order to understand your description in terms of the basis vectors, \(\rhat_P\text{,}\) \(\that_P\text{,}\) and \(\phat_P\text{,}\) at \(P\text{!}\)