Section 21.8 Derivatives of Basis Vectors
Differentiating a vector field expressed in terms of the standard, rectangular basis vectors is easy: Just differentiate the coefficients. Why? Because the rectangular basis vectors are constant, that is
\begin{equation}
d\xhat = d\yhat = d\zhat = 0 .\tag{21.8.1}
\end{equation}
However, if the vector field is expressed in terms of curvilinear basis vectors, the product rule must be used! That is, both the coefficients and the basis vectors must be differentiated. So we need to determine the differentials of the basis vectors.
We start with polar coordinates. One could of course express the polar basis \(\{\rhat,\phat\}\) in terms of the rectangular basis \(\{\xhat,\yhat\}\) as in Section 21.7 and differentiate, then transform the result back to the original, polar basis. But this calculation can also be done entirely in the polar basis, as we show here.
Recall from Section 1.16 that the position vector in polar coordinates is given by
\begin{equation}
\rr = r\,\rhat .\tag{21.8.2}
\end{equation}
Zapping both sides of (21.8.2) with \(d\) and using the product rule yields
\begin{equation}
d\rr = dr\,\rhat + r\,d\rhat .\tag{21.8.3}
\end{equation}
\begin{equation}
d\rhat = d\phi\,\phat .\tag{21.8.4}
\end{equation}
To find \(d\phat\text{,}\) we use the product rule together with the orthonormality of the polar basis. Starting from
\begin{equation}
\phat\cdot\phat = 1\tag{21.8.5}
\end{equation}
we immediately have
\begin{equation}
d\phat\cdot\phat + \phat\cdot d\phat = 2d\phat\cdot\phat = 0 ,\tag{21.8.6}
\end{equation}
a property that holds for any unit vector. Thus, the \(\phat\) component of \(d\phat\) must vanish. Similarly, starting from
\begin{equation}
\phat\cdot\rhat = 0\tag{21.8.7}
\end{equation}
we obtain
\begin{equation}
d\phat\cdot\rhat + \phat\cdot d\rhat = 0\tag{21.8.8}
\end{equation}
or equivalently
\begin{equation}
d\phat\cdot\rhat
= -\phat\cdot d\rhat
= -\phat\cdot(d\phi\,\phat)
= -d\phi .\tag{21.8.9}
\end{equation}
Since we have already shown that the \(\phat\) component of \(d\phat\) vanishes, we can conclude that
\begin{equation}
d\phat = -d\phi\,\rhat .\tag{21.8.10}
\end{equation}
In particular, if we want the time derivatives of the basis vectors, we can divide (21.8.4) and (21.8.10) by \(dt\) to obtain
\begin{align}
\frac{d\rhat}{dt}
\amp = \;\;\,\frac{d\phi}{dt}\, \phat\tag{21.8.11}\\
\frac{d\phat}{dt}
\amp = -\frac{d\phi}{dt}\, \rhat\tag{21.8.12}
\end{align}
Extending this argument to spherical coordinates isn’t quite as straightforward. It’s easy to show that (21.8.4) becomes
\begin{equation}
d\rhat = d\theta\,\that + r\,\sin\theta\,d\phi\,\phat\tag{21.8.13}
\end{equation}
using an argument similar to the one above. But determining the angular derivatives (without using rectangular basis vectors) requires a different approach. The result, which could of course also be obtained using rectangular basis vectors, is
1
One possibility is to use the fact that mixed partial derivatives commute, so that for instance \(\Partials{\rr}{\theta}{\phi} = \Partials{\rr}{\phi}{\theta}\text{,}\) along with the fact that the expression for \(d\rr\) in spherical coordinates (see the spherical coordinates section of Appendix B.2) tells us that \(\Partial{\rr}{\theta}=r\,\that\) and \(\Partial{\rr}{\phi}=r\sin\theta\,\phat\text{.}\) Another approach is to use differential forms (see for instance Chapter 6 of
GDF).
\begin{gather}
d\that = -d\theta\,\rhat + \cos\theta\,d\phi\,\phat ,\tag{21.8.14}\\
d\phat = -\sin\theta\,d\phi\,\rhat - \cos\theta\,d\phi\,\that .\tag{21.8.15}
\end{gather}
