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

Let the angular velocity vector be the spatially constant $\mathbf{\omega }$. Choose an origin on the fixed rotation axis. The velocity field of a rigid body is $v=\mathbf{\omega }\times r$. In components,Taking the curl givesAn added spatially uniform translational velocity has zero curl, so it does not affect this relation. Equivalently, the vorticity of rigid rotation is twice the angular velocity; the local rotational part of a general smooth flow has angular velocity $\frac{1}{2}\nabla \times v$.

For an oriented piecewise smooth surface $S$ with boundary $\partial S$, and a vector field $F$ continuously differentiable on a neighborhood of $S$, Stokes theorem statesHere $n$ is the chosen unit normal vector, $dS$ is the unsigned surface area element, and $dr$ is the tangent displacement along the boundary. The right-hand rule pairs the boundary orientation with $n$. If there are several boundary components, the line integral includes all of them with their induced orientations.

A simply connected surface is connected and every closed loop on it can be continuously contracted to a point while remaining on the surface. A disc is an example. An annulus is not simply connected: a loop winding once around its hole cannot contract within it.

Stokes' theorem is valid for both types of surface, under the smoothness and orientation hypotheses above. Simple connectivity is not a requirement of the theorem. For a concrete annular example, take $F=(-y/2,x/2,0)$ and $a\le r\le R$ in the horizontal plane. Its curl is $(0,0,1)$, so the left side is $\pi (R^2-a^2)$. The right side is the counterclockwise outer-circle integral $\pi R^2$ minus the counterclockwise inner-circle integral $\pi a^2$, giving exactly the same answer. Omitting the inner boundary would be an incorrect application of the theorem.

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.