= Solution
The lower derivative index and the upper <electromagnetic four-potential> index are fully contracted. For a constant <Lorentz transformation>,
$$
\partial'_\alpha A'^\alpha
=(\Lambda^{-1})^\beta{}_{\alpha}\partial_\beta(\Lambda^\alpha{}_{\gamma}A^\gamma)
=\partial_\beta A^\beta.
$$
Hence
$$
\boxed{\partial_\alpha A^\alpha\text{ is a Lorentz-invariant scalar}.}
$$
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.
Back to article page
