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    $\begingroup$ The very definition of a gauge condition is that it is physically as "good" and as "bad" as any other, that is, the choice of gauge condition does not impact the observable physics at all. Why do you think it has a "physical meaning"? $\endgroup$
    – ACuriousMind ♦
    Commented Jul 15, 2016 at 21:54
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    $\begingroup$ The question should be when and for what purposes is it more useful to use this gauge condition other than another, such as under what conditions does it give you calculational advantages or makes it easier to interpret results? $\endgroup$
    – Bob Bee
    Commented Jul 15, 2016 at 22:23
  • $\begingroup$ There is none. There only is a technical convenience of using it in some problems. In terms of potentials there is no "gauge invariant core" and additional "purely gauge terms" to throw away. $\endgroup$
    – Vladimir Kalitvianski
    Commented Jul 16, 2016 at 9:06