= Solution
On the <sphere>, constant $\theta$ gives a latitude <circle> of radius $a\sin\theta$ and height $a\cos\theta$. Constant $\phi$ gives a meridian semicircle in a vertical plane through the $z$ axis. The <orthonormal basis> of <spherical coordinates> is
$$
\boldsymbol e_r=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta),\quad
\boldsymbol e_\theta=(\cos\theta\cos\phi,\cos\theta\sin\phi,-\sin\theta),\quad
\boldsymbol e_\phi=(-\sin\phi,\cos\phi,0).
$$
Here $\boldsymbol e_r$ points outward, $\boldsymbol e_\theta$ along increasing <polar angle>, and $\boldsymbol e_\phi$ along increasing <azimuth>. At either pole the radial <unit vector> is unambiguous, but the two tangential coordinate vectors depend on the undefined longitude. The figure shows their directions away from the poles.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2008/paper-1-spherical-coordinate-basis.png]
{title=Latitude and meridian curves with the three spherical coordinate unit vectors at a regular point}
This coordinate ambiguity does not make a vector field singular if its tangential coefficient vanishes suitably. In <Cartesian coordinates>,
$$
\boldsymbol A^{(+)}=\frac{(-y,x,0)}{r(r+z)},\qquad
\boldsymbol A^{(-)}=\frac{(y,-x,0)}{r(r-z)}.
$$
The first is smooth at the north pole and zero there; the second is smooth and zero at the south pole. Their actual singular axes are the opposite ones. These are the regular patches of a <two-patch vector potential for a radial monopole>.
Back to article page
