0
$\begingroup$

"Vector Potential Coil and Transformer" by M. Diabo et al reports the induction of voltage by a time varying curl free vector potential. A refinement reported in "Vector-Potential Coil and Transformer - With a Superconducting Secondary Coil and Superconducting Magnetic Shield" reproduces this induced voltage within a superconducting magnetic shield.

My understanding is that such an experimental result is inconsistent with gauge invariance, hence Special Relativity.

Is this true?

$\endgroup$
  • $\begingroup$ Why would gauge variance violate special relativity? I can argue the contrary. $\endgroup$ – my2cts Feb 13 at 19:44
  • $\begingroup$ Classical electrodynamics are gauge invariant and special relativity relies on classical electrodynamics. ippp.dur.ac.uk/~krauss/Lectures/QuarksLeptons/QED/… $\endgroup$ – James Bowery Feb 14 at 21:49
  • $\begingroup$ The fact is that one does not imply the other. $\endgroup$ – my2cts Feb 15 at 15:08
  • $\begingroup$ To be clear, by "does not imply" do you mean "special relativity does not rely on classical electrodynamics" or "classical electrodynamics does not rely on gauge invariance" or both? $\endgroup$ – James Bowery Mar 17 at 23:22
  • $\begingroup$ Special relativity does not imply gauge invariance, nor the other way around. $\endgroup$ – my2cts Mar 18 at 0:02
1
$\begingroup$

No, it isn't. The authors measure induced electromotive force due to primary winding to be in agreement with their calculation. It seems in agreement with standard EM theory.

In the calculation, they implicitly use the vector potential in the Coulomb gauge ($\nabla\cdot \mathbf A = 0$), so they can use the well-known solution of the equation $\mathbf j = \nabla\times\mathbf B$ (B as a function of j) to express solution to the equation $\mathbf B = \nabla\times \mathbf A$ (A as a function of B, or flux $\phi$).

Then they assume induced electric field is given accurately by this $\mathbf A$ only:

$$ \mathbf E = -\frac{\partial \mathbf A}{\partial t} $$

although in general, one should include also contribution of the form $-\nabla \varphi$, where $\varphi$ is the corresponding electric potential, which in this gauge is the Coulomb potential of all charges.

Its neglect here is probably a very accurate assumption, as the plus/minus poles of the primary are put very close together, near the AC voltage source, and there is no place for the charges to accumulate and form substantial variation of the Coulomb electric potential.

Also, let me say that this does not mean that observed electric field is a sign of "this coulomb-gauge A being real". They do claim that the measured voltage on the oscilloscope is due to A and not due to magnetic field, but this is misleading or revealing a misunderstanding on their part. No law of electromagnetism requires that magnetic field be present at the same region where induced electric field is present; the magnetic field may be in some other region, next to the one where the voltage is being measured. In this case, there is magnetic field and magnetic flux present - it is confined to the primary coil, and this is all that is needed to make the phenomenon consistent with the Faraday law. So, no new physical effect of vector potential is observed.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.