 Solution

An infinitesimal rotation about $x$ gives $\delta {\psi }_y=-ϵ{\psi }_z$ and $\delta {\psi }_z=ϵ{\psi }_y$. Both the squared transverse velocity and squared transverse slope are rotational invariants. Noether's theorem therefore givesDifferentiating and using the two wave equations givesConsequently $L_x=\int {\ell }_{x}dx$ is the angular momentum about the equilibrium string axis, conserved when the net endpoint torque flux vanishes. For a finite segment, $d{L}_x/dt=-[{\mathcal{J}}_x{]}_{\mathrm{left}}^{\mathrm{right}}$. It is not momentum associated with translating the coordinate $x$; its symmetry rotates the two physical transverse displacement components into one another.