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

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.