According to my current knowledge, the Cooper-pairs in a closed superconducting circuit should "feel" a centrifugal force. The (-2) charged pairs are therefore accelerated.

Do the Cooper pairs emit photons? Is there some mechanism that prevents it? Even if you look at the complete wavefunction, still there is an electric charge that accelerates. At different points in the circuit, this acceleration is different. I can't find any reason why they shouldn't, but I'm not sure. I'm not sure either if this case the form of the circuit has the same implications as that of a closed circular circuit (apparently not, according to the comments below).

Has the solution to this question has something to do with the Meissner effect, as in the answer below (though I can't see how)?
Or is the emission that small (because of the small acceleration due to the low temperature) that it effectively has no effect on the current's energy? Just stating that the definition of a superconductor is that it experiences no resistance begs the question.

So, Is there anyone who can help me solve this riddle? Much obliged in advance!


2 Answers 2


Steady currrents do not radiate, as per Maxwell equations. This has nothing to do with superconductivity. As for why the electrons Bremsstrahlung cancels out, see http://kirkmcd.princeton.edu/examples/steadycurrent.pdf .

  • $\begingroup$ But this is not a steady current. It's radially accelerated (like a ball on a rope traces out a circular trajectory). $\endgroup$ Aug 27, 2020 at 7:23
  • $\begingroup$ @descheleschilder, yes, the charges are radially accelerated. But that doesn't mean that the current is not steady. $\endgroup$
    – BowlOfRed
    Aug 27, 2020 at 7:46
  • $\begingroup$ The argument provided in my reference holds for a closed loop. $\endgroup$ Aug 27, 2020 at 7:49
  • $\begingroup$ @BowlOfRed Well, that's the question. $\endgroup$ Aug 27, 2020 at 7:57
  • 1
    $\begingroup$ @descheleschilder Me too, but I was also taught Maxwell equations. And, for the third time, I provided a reference illustrating how the Bremsstrahlung (one h, not two) cancels. $\endgroup$ Aug 27, 2020 at 8:12

Superconducting currents do not radiate. If they did, then energy and momentum conservation would lead to a slowing down of the current, but you correctly say that does not occur. As for the mechanism that prevents radiation I suspect it is some variation of the Meissner effect.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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