4

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?

David Jonsson
  • 936

1 Answers1

16

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.

Ivan Burbano
  • 3,875
  • 1
    This answer makes the AB effect to be as simple as Stokes's theorem. I guess we need to keep the chosen gauge if measuring at different occasions? – David Jonsson Aug 22 '20 at 10:56
  • 2
    I am sorry, I don't think I understand. If measuring what at different locations? If you mean that one needs to keep the same gauge when computing the integral $\oint_\mathcal{C} A$ then yes. You cannot change your gauge in the middle of the computation of that integral. – Ivan Burbano Aug 22 '20 at 12:22
  • Sorry. Yeah, you should definitely use the same gauge for the computation of the integral throughout the path of the particle. – Ivan Burbano Aug 23 '20 at 07:36
  • I mean that the path will not be closed if the system is moving. It will be a combination of a cycloid and a helix. They could become closed with an extra path. Is there any analysis of that? – David Jonsson Sep 10 '20 at 22:57
  • I don't think I understand. To the best of my understanding, all experiments require some sort of closed path (a particle traveling a closed path, or two particles going from point A to B through different paths, etc....) – Ivan Burbano Sep 10 '20 at 23:43
  • Did you deliberately leave out integration symbols? The AB effect can take any value in space. In spacetime however it does not exist. It is the closed path choice, which is maybe not closed in spacetime or just seemingly closed due to space choise, that can violate relativity in the sense that it may not be closed in spacetime. – David Jonsson Jul 20 '21 at 14:44
  • Oh, I see what you mean. I am sorry I didn't understand before. So, I believe that in the AB effect the electromagnetic field is constant in time. Therefore, the phase gained can be obtained by taking $\mathcal{C}$ to be a loop at a constant time slice (just project onto a timeslice the trajectory of the particle). In more general situations however, taking a general loop $\mathcal{C}$ in spacetime is still physically relevant. For example, the trajectory of two electrons with the same start and end points effectively forms a closed loop in spacetime. – Ivan Burbano Jul 20 '21 at 19:37
  • As for the integration symbols, I wrote everything in terms of differential forms. In components the integrals I wrote look like $\int_\mathcal{C} A_\mu \text{d}x^\mu$ and $\int_\Sigma \frac{1}{2}F_{\mu\nu}\text{d}x^\mu\wedge\text{d}x^\nu$. – Ivan Burbano Jul 20 '21 at 19:39
  • I can also add that in the original AB effect the four-dimensional notation is not necessary. In WP https://en.wikipedia.org/wiki/Aharonov%E2%80%93Bohm_effect#Magnetic_solenoid_effect they describe how to express the phase in terms of the magnetic flux (i.e. in a gauge-invariant way). – Ivan Burbano Jul 20 '21 at 19:42