For the <active Lorentz transformation of a vector field>, coordinates are held fixed and $A'^\mu(x)=\Lambda^\mu{}_{\nu}A^\nu(\Lambda^{-1}x)$. Its field strength transforms as a two-index <Lorentz tensor>. The index contractions in the <Maxwell Lagrangian> are invariant by $\Lambda^T\eta\Lambda=\eta$, so the density transforms as a <Lorentz scalar>:
$$
\mathcal L'(x)=\mathcal L(\Lambda^{-1}x).
$$
For $\Lambda=I+\alpha\omega$, its inverse is $I-\alpha\omega+O(\alpha^2)$. Expanding the coordinate argument therefore gives
$$
\boxed{\delta\mathcal L(x)=-\alpha\omega^\mu{}_{\nu}x^\nu\partial_\mu\mathcal L(x).}
$$
Set $v^\mu(x)=\alpha\omega^\mu{}_{\nu}x^\nu$. Part (iii) implies $\partial_\mu v^\mu=\alpha\omega^\mu{}_{\mu}=0$. The product rule then converts the variation into a total derivative:
$$
\boxed{\delta\mathcal L=-\partial_\mu\bigl(\alpha\omega^\mu{}_{\nu}x^\nu\mathcal L\bigr).}
$$
Its integral changes the action only by a boundary term, which is the condition needed for <Noether's theorem>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 301 / 4 / iv / Solution

An active transformation keeps the coordinate argument fixed and changes a contravariant vector field by $A'^\mu(x)=\Lambda^\mu{}_{\nu}A^\nu(\Lambda^{-1}x)$. Infinitesimally, $\delta A^\mu=\alpha\omega^\mu{}_{\nu}A^\nu-\alpha\omega^\rho{}_{\sigma}x^\sigma\partial_\rho A^\mu$. For a <Lorentz scalar> density, the spin-index term is absent, giving $\delta\mathcal L=-\alpha\omega^\rho{}_{\sigma}x^\sigma\partial_\rho\mathcal L$.

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 Incoming links: Active Lorentz transformation of a vector field