Past exam of the mathematical methods II course of the University of Oxford / 2019 / 8 / c

Write $r^2=x^2+y^2$ and work away from the $z$ axis, where $r>0$. The vector field is independent of $z$ and has zero vertical component, so only the vertical curl can be nonzero:ThereforeThis local spin in an azimuthal power-law flow depends on the radial variation of the tangential speed, not merely on the sign of that speed.

In cylindrical coordinates, $F=r^{1-2n}\hat{\theta }$. The tangential coefficient is positive for every real $n$ and $r>0$, so the large-scale circulation is always counterclockwise when viewed from positive $z$.

By solution, the infinitesimal local angular velocity is $(1-n)r^{-2n}\hat{z}$. Its sign gives“Zero local spin” does not mean zero velocity: at $n=1$ a marked small element is still carried around the axis while its infinitesimal rotational component is zero. Orbital motion around the axis and local vorticity are different measurements.

For $n=1$, the azimuthal unit vector is $\hat{\theta }=(-\sin \theta ,\cos \theta ,0)$, soOn a counterclockwise circle $C_r$ at fixed height, $dr=r\hat{\theta }d\theta$. Henceindependent of the positive radius. Clockwise orientation reverses the sign.

There is no contradiction with the zero curl found in solution. The vector field is undefined on the axis, so it is not continuously differentiable on a disc spanning $C_r$; the hypotheses of Stokes theorem fail for that disc. It is smooth on a planar annulus excluding the axis, where Stokes theorem correctly gives zero: the two boundary line integrals cancel, $2\pi -2\pi =0$.

The same check works for general $n$. A counterclockwise circle has circulation $\Gamma (r)=2\pi r^{2-2n}$. Integrating the curl over an annulus $a\le r\le R$ gives, for $n\ne 1$,For $n=1$, both sides are zero. Thus all three preceding results are consistent with Stokes theorem; the necessary distinctions are the regularity of the vector field and inclusion of every oriented boundary component.