Consider the transformation matrix from cartesian $(x,y)$ to polar $(R,\phi)$ coordinates:
\begin{equation}\left[ \frac{\partial x'^a}{\partial x^b} \right]=
\begin{bmatrix}
\cos(\phi) & \sin(\phi) \\
-\frac{1}{R}\sin(\phi) & \frac{1}{R}\cos(\phi)
\end{bmatrix}
\end{equation}
where the primed coordinates are the polar coordinates.
To transform the unit vector $\hat e_x=(1,0)$ to polar coordinates, using $x'^a=\frac{\partial x'^a}{\partial x^b}x^b$, will lead me to:
$\hat e_x=\cos(\phi)\hat e_R-\frac{1}{R}\sin(\phi)\hat e_\phi$, which isn't correct, due to the $\frac{1}{R}$. Where did I go wrong?
Best Answer
It's not wrong. It's totally correct. Your choice of $\hat{e}_{\phi}$ (implicit in the Jacobian, as you used the function $\phi$ and not the function $R\phi$) makes its norm proportional to $R$, hence the normalising coefficient $\frac{1}{R}$.