Dilatation current ($D^{\mu }$)

The dilatation current associated with scale transformations is $D^{\mu }=x_{\nu }{T}^{\mu \nu }$ up to improvement terms. Its divergence is the trace of the stress-energy tensor, so classical scale invariance requires a traceless improved tensor on shell.

For a first-derivative real scalar field in one space and one time dimension, put $j_D^{\mu }={\mathcal{L}}_{{\partial }_{\mu }\psi }(z{\psi }^{\prime }+t\dot{\psi })-x^{\mu }\mathcal{L}$. Its on-shell divergence isA quadratic derivative density with arbitrary field-dependent coefficients has zero derivative-homogeneity term. An extra term $-\alpha {\psi }^{\prime 4}$ instead gives $-2\alpha {\psi }^{\prime 4}$. A position-dependent kinetic coefficient $A\rho (z)/2$ gives $-Az{\rho }^{\prime }(z){\dot{\psi }}^2/2$. These source terms distinguish scale invariance from internal field-shift symmetry.

For constant $A,\rho ,K$ on all space and time, the active waveform transformation ${\psi }_b(z,t)=\psi (bz,bt)$ leaves the action of $\mathcal{L}=A(\rho {\dot{\psi }}^2-K{\psi }^{\prime 2})/2$ invariant: the density gains $b^2$ and the integration measure gains $b^{-2}$. With canonical momentum $\Pi =A\rho \dot{\psi }$, its Noether current can be chosen asThe integrated charge $D=\int j_D^{0}dz$ is conserved if its boundary flux vanishes and the weighted integral exists. For a rightgoing wave packet $\psi =f(z-ct)$, $c=\sqrt{K/\rho }$, it becomes $D=-A\rho c\int \xi [f^{\prime }(\xi ){]}^{2}d\xi$. A fixed finite interval generally breaks this scale symmetry unless its boundaries and boundary data are transformed too.