Past exam of the optics and electrodynamics course of the University of Cambridge / 2021 / Section B / B6 / a

The upper index identifies $P^{\alpha }$ as a contravariant Lorentz four-vector, obeying the rank-one contravariant tensor transformation law. Its components transform with the coordinate Jacobian,It is not invariant component by component. Its contracted square $P_{\alpha }{P}^{\alpha }=m^2{c}^2$, for metric signature $(+,-,-,-)$, is a Lorentz invariant. This distinguishes a contravariant transformation law from an invariant scalar.

First lower the index of the four-current with the Minkowski metric: $J_{\beta }=g_{\beta \alpha }{J}^{\alpha }$. The displayed expression is thenThereforeIts one free lower index transforms with the inverse coordinate Jacobian. The contraction with the metric here lowers an index; it does not make the current a scalar because the index $\delta$ remains free.

The chain rule givesThusDifferentiation with respect to a contravariant coordinate has a lower transformation index. In inertial Cartesian coordinates in Minkowski spacetime, no connection term is needed for this coordinate transformation of a derivative acting on a scalar; Lorentz transformations have constant Jacobians.

The lower derivative index and the upper electromagnetic four-potential index are fully contracted. For a constant Lorentz transformation,HenceThis is a statement about Lorentz transformations, not electromagnetic gauge transformations. The quantity is not generally gauge invariant; setting it to zero is the Lorenz gauge condition, which is Lorentz covariant because zero remains zero in every inertial frame.