= 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 $X=z\psi'+t\dot\psi$. Since the density is quadratic in derivatives,
$$
\frac{\delta\mathcal L}{\eta}=2\mathcal L+z\partial_z\mathcal L+t\partial_t\mathcal L
=\partial_t(t\mathcal L)+\partial_z(z\mathcal L).
$$
The variation is a divergence, so subtract its boundary term in the <Noether current>:
$$
j_2^0=\Pi X-t\mathcal L=t\mathcal H+z\Pi\psi',\qquad
j_2^z=-AK\psi'X-z\mathcal L=-tAK\dot\psi\psi'-z\mathcal H.
$$
Thus the <dilatation charge of a one-dimensional elastic wave> is
$$
\boxed{Q_2=\int_{-\infty}^{\infty}\left(t\mathcal H+z\Pi\psi'\right)dz}.
$$
An overall sign change would give the same <conservation law>. Unlike $Q_1$, this charge has units of <action>, $\mathrm{J\,s}$. Its explicit factor of $t$ is balanced by transport of the weighted <momentum> density; an explicitly time-dependent charge can still be conserved.
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