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Gauge condition can be chosen as you like or not? is the Lorenz gauge is the only one correct? If Coulomb gauge can obtained exactly same results as Lorenz gauge for the electromagnetic fields E and B, we can choose what we like, either Coulomb gauge or Lorenz gauge. If use different gauge we get different formula of electromagnetic fields $E$ and $B$, then we have to choose a corrected gauge to get the correct electromagnetic field formula. I am not clear that when we use Lorenz gauge or Coulomb gauge, the result formula for electric $E$ and magnetic field $B$ same or different?

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Gauge condition is any human imposed restriction on the functions $\varphi(\mathbf x,t), \mathbf A(\mathbf x,t)$ that does not change electric and magnetic field implied by those potential functions.

For any single physical situation, one can use either potential functions obeying the Coulomb gauge condition or those obeying the Lorenz gauge condition. Both choices give the same electric and magnetic field.

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  • $\begingroup$ Aharonov–Bohm effect proves that the vector potential A and scale potential $**\phi**$ has more physics meaning the electromagnetic field E and B. I have the felling if A and \phi is obtained by using different gauge, the corresponding electric field and magnetic field are also different. Could someone prove that the electric field and magnetic field from different gauges are same? $\endgroup$ – ShRenZhao Jan 19 at 0:38
  • $\begingroup$ The Aharonov-Bohm shift was measured to exist and be close to their prediction, but it is not clear those experiments prove the Aharonov-Bohms explanation of the effect that is based on superiority of the Coulomb-gauged potential as compared to EM field. There are other ways to explain Aharonov-Bohm shift that use only local interaction with EM field. See papers by Timothy Boyer on this subject, e.g. ui.adsabs.harvard.edu/abs/1973PhRvD...8.1679B/abstract $\endgroup$ – Ján Lalinský Jan 19 at 0:52

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