= Shift current of an elastic rod
For a longitudinal displacement field $\psi(z,t)$ and <Lagrangian density> $\mathcal L=A[\rho\dot\psi^2-K\psi'^2]/2$, uniform field shifts give <Noether current> $j^0=A\rho\dot\psi$, $j^z=-AK\psi'$. Its spatial integral is the rod's longitudinal mechanical <momentum>. A prescribed position-dependent <mass density> changes $j^0$ to $A\rho(z)\dot\psi$ but does not break this internal shift symmetry. Boundary traction can carry momentum into or out of a finite rod, so action symmetry alone does not guarantee conservation of the integrated charge.
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