Transverse momentum current of an elastic string
= Transverse momentum current of an elastic string
{title2=$j_y^\mu=(\rho\dot\psi_y,-F\psi_y\prime)$}
For $\mathcal L=\rho(\dot\psi_y^2+\dot\psi_z^2)/2-F((\psi_y')^2+(\psi_z')^2)/2$, constant displacement symmetry in $y$ gives $j^0=\rho\dot\psi_y$ and $j^x=-F\psi_y'$. The <wave equation> implies $\partial_tj^0+\partial_xj^x=0$. Its integrated charge is physical transverse momentum; endpoint forces can change that charge on a finite segment.