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## Main Question or Discussion Point

I am trying to show that the four-potential is a four-vector when working in the Lorenz gauge. In this gauge, we have ## \Box A^{\mu} = 4\pi J^{\mu}##. If we perform a Lorentz transformation, we can show that ## \Box A'^{\mu} = \Box \Lambda_{\nu}^{\mu} A^{\nu}##. From what I have seen, people have used this to conclude that ## A'^{\mu} = \Lambda_{\nu}^{\mu} A^{\nu}##, but I don't see why this is necessarily the case. If we had any function ## \xi ## such that ## \Box \xi =0 ##, then we could have ## A'^{\mu} = \Lambda_{\nu}^{\mu} A^{\nu} + \partial^{\mu} \xi## instead, where ##\partial^{\mu} \xi## is not necessarily zero.

In a previous thread, someone invoked the quotient theorem. However the quotient theorem describes situations like if ## B_{\mu} C^{\mu \nu}## is a tensor whenever ## B_{\mu} ## is a tensor, then ##C^{\mu \nu}## is a tensor. This does not apply to our case, as we only know that ## S A^{\mu} ## is a tensor only when ## S = \Box##; we do not have the universal quantification necessary to invoke the quotient theorem.

In a previous thread, someone invoked the quotient theorem. However the quotient theorem describes situations like if ## B_{\mu} C^{\mu \nu}## is a tensor whenever ## B_{\mu} ## is a tensor, then ##C^{\mu \nu}## is a tensor. This does not apply to our case, as we only know that ## S A^{\mu} ## is a tensor only when ## S = \Box##; we do not have the universal quantification necessary to invoke the quotient theorem.