Past exam of the theoretical physics I course of the University of Cambridge / 2017 / 1 / b

Both Noether charges require boundary conditions that admit the stated transformations and no net boundary-current flux. Take a sufficiently localized field on the whole rod, with the weighted integrals below finite. Fixed endpoints at finite positions generally spoil the waveform-dilation symmetry.

The transformation $\psi \mapsto \psi +b$ leaves both derivatives unchanged, so the Lagrangian density and action are invariant. For infinitesimal unit field variation, the Noether current for a first-derivative scalar field is the shift current of an elastic rod,Its divergence vanishes by the equation of motion. The charge is the total longitudinal mechanical momentum to this order in the displacement field.

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.