 Solution

ID: past-exam-of-the-optics-and-electrodynamics-course-of-the-university-of-cambridge/2021/section-b/b6/a/iv/solution

The lower derivative index and the upper electromagnetic four-potential index are fully contracted. For a constant Lorentz transformation,

\partial_{α}^{'} A^{' α} = (Λ^{- 1})^{β}_{α} \partial_{β} (Λ^{α}_{γ} A^{γ}) = \partial_{β} A^{β} .

Hence

\partial_{α} A^{α} is a Lorentz-invariant scalar .

This 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.

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