 Shift current of an elastic rod

For a longitudinal displacement field $\psi (z,t)$ and Lagrangian density $\mathcal{L}=A[\rho {\dot{\psi }}^2-K{\psi }^{\prime 2}]/2$, uniform field shifts give Noether current $j^0=A\rho \dot{\psi }$, $j^z=-AK{\psi }^{\prime }$. 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.