The Lorenz gauge is the only Lorentz invariant electrodynamic gauge. If the vector potential has physical meaning, as in the Aharonov-Bohm effect (ABE), then the gauge condition can not be arbitrarily chosen and Lorentz invariance seems gone.
How can the integration path in ABE be Lorentz invariant?
This is not the case. The Aharanov-Bohm effect yields an observable of the form $$\oint_{\mathcal C} A,$$ where $\mathcal{C}$ is some circuit. This is however gauge invariant. A way to see this is by noting that it can be written in terms of the field strength $$\int_\Sigma F,$$ via Stokes' theorem. In here we chose a surface $\Sigma$ whose boundary is $\mathcal{C}$. The point however is that in the Aharanov-Bohm effect, the observable corresponds to an experiment that happens in $\mathcal{C}$, not $\Sigma$. In particular, the unintuitive part of it is that there is no apparent reason why the observable should contain information of $F$ everywhere on $\Sigma$ if the particles never went there. Therefore, if you want to describe physics locally, you must use the potential $A$ instead of the field strength $F$. Gauge invariance is however not lost.
