Past exam of the mathematics course of the University of Cambridge / 2003 / ia / Paper 3 / 11A (Vector Calculus)

Use suffix notation, summing repeated indices. The contraction of two Levi-Civita symbols givesAdding the directional derivative $(F\cdot \nabla )G_i=F_j{\partial }_j{G}_i$ cancels the second term. Interchanging $F,G$ gives another pair whose sum is $G_j{\partial }_i{F}_j$. The total right-hand side is thereforethe $i$th component of the gradient of the dot product. Since this holds for each component, the dot-product gradient identity is proved:It holds for any continuously differentiable vector fields; no irrotationality assumption is used.

Take $E$ to be continuously differentiable everywhere on ${\mathbb{R}}^3$, as in the question. Its zero curl says ${\partial }_i{E}_j={\partial }_j{E}_i$. Construct a potential of a conservative vector field explicitly byAll segments lie in ${\mathbb{R}}^3$. Differentiating under the integral and using the symmetry of the derivatives,Hence $E=-\nabla \varphi$. This radial integral potential for a curl-free field also works on a star-shaped domain after translating its centre. On an arbitrary domain, zero curl only guarantees local potentials; the everywhere-on-${\mathbb{R}}^3$ interpretation is important for this global assertion.

For the given field, put $q=e^{-x^{2}z}$. The three pairs of derivatives entering its curl areThus the field is irrotational. To find its potential using $E_y=-{\partial }_y\varphi$, integrate ${\partial }_y\varphi =2yq$ to get $\varphi =y^{2}q+\psi (x,z)$. Comparison with $E_x=-{\partial }_x\varphi$ gives ${\psi }_x=0$, and comparison with $E_z=-{\partial }_z\varphi$ gives ${\psi }_z=0$. ThereforeDifferentiating this expression reproduces all three components with the required minus sign; the only freedom on the connected domain is the additive constant.