Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2016 / Paper 1 / 20 / a

On a connected domain, a vector field is a conservative vector field if and only if it is the gradient of a single-valued scalar potential, $\boxed{\mathbf{H}=\nabla \varphi }$. Equivalently, its line integral is path independent, or its integral around every closed curve is zero. For a continuously differentiable field on a simply connected domain, this is equivalent to $\nabla \times \mathbf{H}=0$. Without the domain hypothesis, vanishing curl alone is only a local test and is not the necessary-and-sufficient global statement.

Yes. For any sufficiently smooth scalar $\chi$, the gauge transformationleaves its curl unchanged because $\nabla \times \nabla \chi =0$. This freedom is much larger than adding a constant vector: for example, $\chi =xy$ adds the nonconstant vector $(y,x,0)$. Two vector potentials for the same field differ by a curl-free field; on a simply connected domain that difference is a gradient, while nontrivial topology may allow additional global freedom.

Take the curl component by component:Its divergence is zero, as expected because the divergence of a curl is zero.