Past exam of the mathematical methods II course of the University of Oxford / 2014 resit / 11

Write $\mathbf{\omega }=({\omega }_x,{\omega }_y,{\omega }_z)$. The cross product gives $\mathbf{v}=({\omega }_{y}z-{\omega }_{z}y,{\omega }_{z}x-{\omega }_{x}z,{\omega }_{x}y-{\omega }_{y}x)$. Since the components of $\mathbf{\omega }$ are constant, direct partial differentiation givesThusThe vorticity is twice the local angular velocity of this rigid-body rotation.

This is the velocity field of solid-body rotation about the axis parallel to $\mathbf{\omega }$ through the origin. A fluid-following body's centre moves around that axis at angular speed $|\mathbf{\omega }|$, keeping its axial coordinate and perpendicular distance fixed. Its orientation rotates at the same angular velocity. The symmetric part of the velocity-gradient tensor is zero, so the prescribed flow produces no local strain or change of shape. Indeed, the separation $\mathbf{d}$ between two fluid points obeys $d|\mathbf{d}{|}^2/dt=2\mathbf{d}\cdot (\mathbf{\omega }\times \mathbf{d})=0$.

This is a kinematic prediction for an entrained body or material object. A finite inertial body initially at rest cannot instantaneously acquire the prescribed velocity and rotation without forces and torques; its transient after placement would also require density, drag and coupling data. Neglecting gravity alone does not specify that transient. On the rotation axis its centre can remain fixed while its orientation still rotates.

Take the curl component by component:The field is irrotational exactly when all three constant components vanish:

With those constants, integrate ${\varphi }_x=x+3y+3z$ with respect to $x$ to obtain $\varphi =x^2/2+3xy+3xz+g(y,z)$. Matching ${\varphi }_y=3x+y-z$ gives $g=y^2/2-yz+h(z)$. Finally, matching ${\varphi }_z=3x-y-2z$ gives $h=-z^2+C$. Thus a potential of a conservative vector field isThis explicit polynomial establishes a global potential on all of ${\mathbb{R}}^3$; no topological obstruction is present. Its additive constant is arbitrary.

The only contributions to the Laplacian come from the pure quadratic terms. HenceThe scalar potential is therefore a harmonic function. Since $\mathbf{F}=\nabla \varphi$,Thus this field is both irrotational and divergence-free. By the divergence theorem, its net flux through every sufficiently regular closed surface is zero.