Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2005 / Paper 2 / 6

Under an orthogonal Cartesian change $x_i^{\prime }=R_{ij}{x}_j$, the chain rule gives ${\partial }_i^{\prime }=(\partial x_j/\partial x_i^{\prime }){\partial }_j=R_{ij}{\partial }_j$. Thus the derivative operator transforms as a rank-one Cartesian tensor. Under a general coordinate change it has the covector transformation law when applied to a scalar; derivatives of vector components then need the usual connection terms. The calculations below use Cartesian coordinates.

Use $w_j={ϵ}_{j\ell m}{\partial }_{\ell }{u}_m$ and contract the Levi-Civita symbols:This proves the convective acceleration identity, with the sign set by the displayed order $\mathbf{w}\times \mathbf{u}$.

For the curl of a cross product, the Cartesian contraction isThe product rule then givesFor the fluid equation, replace the advective term by $\mathbf{w}\times \mathbf{u}+\nabla (|\mathbf{u}{|}^2/2)$ using the preceding identity. Take its curl. Both gradient terms have zero curl, while $\nabla \cdot \mathbf{w}=\nabla \cdot (\nabla \times \mathbf{u})=0$ and incompressibility gives $\nabla \cdot \mathbf{u}=0$. The vorticity equation is consequentlyThe left side is the material derivative of vorticity; the right is vortex stretching. It follows here because the given force per mass is a gradient, in addition to the flow being inviscid and incompressible.