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Sep 2, 2018 at 13:50 vote accept grodta
Jul 21, 2016 at 12:55 vote accept grodta
Sep 2, 2018 at 13:50
Jul 18, 2016 at 2:17 vote accept grodta
Jul 21, 2016 at 12:55
Jul 18, 2016 at 2:17 vote accept grodta
Jul 18, 2016 at 2:17
Jul 16, 2016 at 9:35 history closed rob
Diracology
Gert
knzhou
ACuriousMind
Needs details or clarity
Jul 16, 2016 at 9:06 comment added Vladimir Kalitvianski 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.
Jul 16, 2016 at 8:12 answer added Selene Routley timeline score: 14
Jul 16, 2016 at 2:24 answer added Timaeus timeline score: 5
Jul 15, 2016 at 22:23 comment added Bob Bee 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?
Jul 15, 2016 at 21:54 comment added ACuriousMind 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"?
Jul 15, 2016 at 21:48 history edited Qmechanic CC BY-SA 3.0
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Jul 15, 2016 at 21:46 review Close votes
Jul 16, 2016 at 9:35
Jul 15, 2016 at 20:29 history edited Qmechanic CC BY-SA 3.0
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Jul 15, 2016 at 20:28 review First posts
Jul 15, 2016 at 21:31
Jul 15, 2016 at 20:25 history asked grodta CC BY-SA 3.0