= Forward displacement from a translating localized potential
{title2=$v\Delta x=\int\dot x^2dt$}
For a translating <potential energy> profile with equal values at the two ends and <linear drag> <force> $-\zeta\dot x$, a finite passing trajectory satisfies $v\Delta x=\int\dot x^2dt\ge0$. Equivalently, since $\int F\,dy=0$, $\Delta x=\int u^2/[v(v-u)]\,dy$. <Compact support> of the <force> alone is insufficient. At high speed the displacement is $v^{-2}\int(F/\zeta)^2dy+O(v^{-3})$.
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