 Solution

Consider an unforced incompressible flow with localized vorticity and sufficient smoothness for integration by parts. Define the hydrodynamic impulse per unit density by $\mathbf{I}=\frac{1}{2}\int \mathbf{x}\times \mathbf{\omega }dV$, denoting the vector impulse by $\mathbf{I}$ to distinguish it from the later scalar $L$. The vorticity equation can be writtenThe supplied identity, integrated over all space, gives $\frac{1}{2}\int \mathbf{x}\times \nabla \times \mathbf{F}dV=\int \mathbf{F}dV$, because $\mathbf{F}$ is localized and its surface terms vanish. ThereforeThe second integral is a vanishing surface integral. The vector identity $\mathbf{u}\times \mathbf{\omega }=\nabla (|\mathbf{u}{|}^2/2)-(\mathbf{u}\cdot \nabla )\mathbf{u}$ and incompressibility reduce the first to surface terms proportional to the quadratic velocity. Those vanish for $\mathbf{u}=O(R^{-3})$. Thus

For a large spherical control volume define ${\mathbf{P}}_R={\int }_{V_R}\mathbf{u}dV$. The impulse relation is $\mathbf{I}=(3/2){\lim }_{R\to \infty }{\mathbf{P}}_R$ for this localized field. The factor is important: impulse is not simply the integral of velocity, since an $R^{-3}$ velocity tail has a nonvanishing contribution in the impulse integration by parts involving a factor of $R$.

Using pressure per unit density $\Pi$ and the Navier-Stokes equation givesThe advective flux is $O(R^{-4})$, the pressure force is $O(R^{-1})$, and the viscous term is $O(\nu R^{-2})$ under the regular far-field derivative decay. All vanish as $R\to \infty$. The localized eddy cannot change its total impulse without a net external momentum input. Pressure redistributes momentum internally but its resultant force on an increasingly distant control surface tends to zero. At finite $R$, boundary forces need not vanish exactly; the all-space limit is the invariant statement.