Past exam of the mathematics course of the University of Cambridge / 2009 / ia / Paper 3 / 3B (Vector Calculus)

A continuously differentiable vector field is an irrotational vector field when its curl vanishes:For the global-potential construction, assume that the domain is connected and simply connected, or more generally that all closed-curve line integrals of $\mathbf{F}$ vanish. Then define the potential of a conservative vector field byTwo paths with the same endpoints differ by a closed curve. On a simply connected domain that curve can be spanned, or contracted through piecewise smooth spanning surfaces, inside the domain. Stokes theorem gives zero circulation because the curl is zero. The value of $V$ is therefore path-independent and $V({\mathbf{x}}_0)=0$. To compute a derivative, append a short coordinate segment to a path ending at $\mathbf{x}$. Then $V(\mathbf{x}+h{\mathbf{e}}_i)-V(\mathbf{x})=-{\int }_0^h{F}_i(\mathbf{x}+s{\mathbf{e}}_i)ds$, so ${\partial }_{i}V=-F_i$.

The domain qualification is necessary and is implicit in the potential request. Curl-free alone on an arbitrary domain does not suffice: $\mathbf{F}=(-y,x,0)/(x^2+y^2)$ on the complement of the $z$-axis has zero curl, but its circulation around a unit circle is $2\pi$. It has no single-valued global potential. The intended construction is valid on all of ${\mathbb{R}}^3$ for a smooth curl-free field, and on any other domain satisfying the condition above.

For the given spherical-coordinate field, the angular components have $r{F}_{\theta }=\cos \theta \cos \varphi$ and $r{F}_{\varphi }=m\sin \varphi$, independent of $r$, so the $\theta$ and $\varphi$ curl components vanish. The radial component isIt vanishes identically exactly when $m=-1$. In that case integrate $F_{\theta }=-r^{-1}{V}_{\theta }$ to obtain $V=-\sin \theta \cos \varphi +C(\varphi )$. The azimuthal equation $F_{\varphi }=-(r\sin \theta {)}^{-1}{V}_{\varphi }$ forces $C^{\prime }=0$, and $F_r=0$ removes any radial dependence. At the given north-pole point the first term is zero, so normalization fixes the constant to zero:The Cartesian expression shows that this potential and the field $\mathbf{F}=\nabla (x/r)$ are smooth for $r>0$, including the spherical-coordinate poles. The origin is excluded because the specified field is singular there; that exclusion does not obstruct the displayed global potential.