 Waveform dilation

For $b>0$, put ${\psi }_b(z,t)=\psi (bz,bt)$. Each derivative gains a factor $b$, so ${\mathcal{L}}_b(z,t)=b^2\mathcal{L}(bz,bt)$. Changing integration variables to $(Z,T)=(bz,bt)$ contributes $b^{-2}$, proving exact invariance of the full-space action.

Set $b=1+\eta$. At fixed coordinates, $\delta \psi =\eta X$ with $X=z{\psi }^{\prime }+t\dot{\psi }$. Since the density is quadratic in derivatives,The variation is a divergence, so subtract its boundary term in the Noether current:Thus the dilatation charge of a one-dimensional elastic wave isAn overall sign change would give the same conservation law. Unlike $Q_1$, this charge has units of action, $\mathrm{J}\mathrm{s}$. Its explicit factor of $t$ is balanced by transport of the weighted momentum density; an explicitly time-dependent charge can still be conserved.