= Vorticity of a linear shear flow
{title2=$\omega_z=-\gamma,\quad\Omega_{{\rm spin},z}=-\gamma/2$}
Use a <linear shear flow> to see why straight <pathlines> can still carry local spin. Faster layers pull past slower ones, and a small symmetric paddle can turn clockwise while its center travels straight ahead.
\Image[/shear-flow-vorticity.png]
{title=Straight particle paths in linear shear coexist with clockwise local spin and material-line deformation}
For the planar <simple shear flow> $\mathbf u=(\gamma y,0,0)$ with constant signed <shear rate> $\gamma$, direct <differentiation> gives
$$
\omega_z=\partial_xu_y-\partial_yu_x=-\gamma,\qquad\Omega_{{\rm spin},z}=-\frac\gamma2.
$$
With $\gamma=1\ \mathrm{s^{-1}}$, these are $-1\ \mathrm{s^{-1}}$ and $-0.5\ \mathrm{s^{-1}}$: both signs indicate clockwise orientation under the <right-hand rule>. Yet a particle initially at $(x_0,y_0)$ follows the straight <pathline> $x(t)=x_0+\gamma y_0t$, $y(t)=y_0$. The <curl> measures differences of <velocity> between neighboring particles, rather than bending of one particle's <pathline>.
The <velocity gradient> has $L_{xy}=\gamma$ as its only nonzero entry. Its <spin tensor> supplies clockwise rotation, while its <rate-of-strain tensor> supplies simultaneous deformation. For a <material line> making angle $\alpha$ to the horizontal, the equation $\dot{\delta\mathbf x}=L\delta\mathbf x$ gives $\dot\alpha=-\gamma\sin^2\alpha$. Initially horizontal and vertical <material lines> therefore have <angular velocities> $0$ and $-\gamma$; their mean is $-\gamma/2$. At $t=0.5\ \mathrm{s}$ with $\gamma=1\ \mathrm{s^{-1}}$, an initially vertical displacement becomes $(0.5,1)$ times its original length and has angle $\arctan 2\approx63.435^\circ$, a clockwise change of $26.565^\circ$. The plotted cross deforms; an actual paddle's response also depends on its shape and interaction with the fluid. The exact local spin statement concerns the <spin tensor>, rather than every tracer's orientation.
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