Past exam of the fluids and waves course of the University of Oxford / 2017 / 2 / a

Orient the planar material curve $C(t)$ counterclockwise, and let $\mathbf{t}$ and $\mathbf{n}$ be its unit tangent and outward unit normal vector. The circulation and outward volume flux areFor $\mathbf{u}=(u,v)$ and this orientation, $\mathbf{t}ds=(dx,dy)$ and $\mathbf{n}ds=(dy,-dx)$, givingHere $M$ is the instantaneous flux of the velocity field, rather than transport relative to the moving material curve; that relative transport is zero. If the enclosed region contains no source and the velocity field is smooth there, incompressibility and the planar divergence theorem imply $M=0$.

For smooth inviscid flow of constant mass density under a conservative body force, Kelvin's circulation theorem states that the circulation around any closed material curve is conserved:The constant-mass density assumption is the usual homogeneous incompressible flow setting here. Merely requiring $\nabla \cdot \mathbf{u}=0$ in a fluid of varying mass density would not by itself make the pressure term below a gradient.

Parameterize the material curve by a periodic particle label $s\in [0,1]$, writing ${\mathbf{r}}_t(s,t)=\mathbf{u}(\mathbf{r}(s,t),t)$. Differentiate its line integral, keeping $s$ fixed:The second term vanishes by periodicity. If the conservative body force per unit mass is $-\nabla \Omega$, the Euler momentum equation givesIts line integral around a closed curve is zero, proving the claim. The velocity field must remain smooth on the moving curve, which cannot pass through a line vortex singularity during this argument. No assumption of irrotational flow was needed.

Define the line vortex strength $\kappa$ by $\mathbf{u}=\kappa {\mathbf{e}}_{\theta }/(2\pi r)$, with $\kappa >0$ meaning counterclockwise motion. On an enclosing circle of radius $R$,because the velocity is tangent to the circle. For any simple positively oriented enclosing curve, the same results follow by Stokes theorem and the divergence theorem on the region between that curve and a small inner circle. Both curl and divergence vanish throughout that intervening annulus. ConsequentlyEquivalently, $\Gamma =(\kappa /(2\pi ))\oint d\theta$ and the total angle change is $2\pi$. If strength is instead denoted by $K$ in $u_{\theta }=K/r$, the answer is $\Gamma =2\pi K$. An enclosing disk through the singular origin cannot be used to infer zero circulation from zero local vorticity.